The document summarizes geometric and harmonic series. It defines geometric series as having the form ΣCrk where C and r are fixed numbers, and each term is r times the previous term. It provides a theorem showing that a geometric series converges to 1/(1-r) if |r|<1 and diverges if |r|≥1. The harmonic series Σ1/k, where each term is the reciprocal of the term number, is shown to diverge using a grouping argument where the partial sums exceed n/2.
Review of Trigonometry for Calculus “Trigon” =triangle +“metry”=measurement =...KyungKoh2
Review of Trigonometry for Calculus “Trigon” =triangle +“metry”=measurement =Trigonometry so Trigonometry got its name as the science of measuring triangles.
Review for the Third Midterm of Math 150 B 11242014Probl.docxjoellemurphey
Review for the Third Midterm of Math 150 B 11/24/2014
Problem 1
Recall that 1
1−x =
∑∞
n=0 x
n for |x| < 1.
Find a power series representation for the following functions and state the radius of
convergence for the power series
a) f(x) = x
2
(1+x)2
.
b) f(x) = 2
1+4x2
.
c) f(x) = x
4
2−x.
d) f(x) = x
1+x2
.
e) f(x) = 1
6+x
.
f) f(x) = x
2
27−x3 .
Problem 2
Find a Taylor series with a = 0 for the given function and state the radius of conver-
gence. You may use either the direct method (definition of a Taylor series) or known
series.
a) f(x) = ln(1 + x)
b) f(x) = sin x
x
c) f(x) = x sin(3x).
Problem 3
Find the radius of convergence and interval of convergence for the series
∑∞
n=1
(x+2)n
n4n
.
Ans. Radius r=2,
√
2 − 2 < x <
√
2 + 2. Problem 4
Find the interval of convergence of the following power series. You must justify your
answers.
∑∞
n=0
n2(x+4)n
23n
.
Ans. −12 < x < 4.
Problem 5
For the function f(x) = 1/
√
x, find the fourth order Taylor polynomial with a=1.
Problem 6
A curve has the parametric equations
x = cos t, y = 1 + sin t, 0 ≤ t ≤ 2π
a) Find dy
dx
when t = π
4
.
b) Find the equation of the line tangent to the curve at t = π/4. Write it in y = mx+b
form.
c) Eliminate the parameter t to find a cartesian (x, y) equation of the curve.
d) Using (c), or otherwise identify the curve.
Problem 7
State whether the given series converges or diverges
a)
∑∞
n=0 (−1)
n+1 n22n
n!
.
b)
∑∞
n=0
n(−3)n
4n−1
.
c)
∑∞
n=1
sin n
2n2+n
.
Problem 8
1
Approximate the value of the integral
∫ 1
0
e−x
2
dx with an error no greater than 5×10−4.
Ans.
∫ 1
0
e−x
2
dx = 1 − 1
3
+ 1
5.2!
− 1
7.3!
+ ... +
(−1)n
(2n+1)n!
+ .... n ≥ 5,
for n=5
∫ 1
0
e−x
2
dx ≈ 1 − 1
3
+ 1
5.2!
− 1
7.3!
+ 1
9.4!
− 1
11.5!
≈ 0.747.
Problem 9
Find the radius of convergence for the series
∑∞
n=1
nn(x−2)2n
n!
.
Ans. R = 1√
e
.
Problem 10
Let f(x) =
∑∞
n=0
(x−1)n
n2+1
.
a) Calculate the domain of f.
b) Calculate f ′(x).
c) Calculate the domain of f ′.
Problem 11
Let f(x) =
∑∞
n=0
cos n
n!
xn.
a) Calculate the domain of f.
b) Calculate f ′(x).
c) Calculate
∫
f(x)dx.
Problem 12
Using properties of series, known Maclaurin expansions of familiar functions and their
arithmetic, calculate Maclaurin series for the following.
a) ex
2
b) sin 2x
c)
∫
x5 sin xdx
d) cos x−1
x2
e)
d((x+1) tan−1(x))
dx
Problem 13
Calculate the Taylor polynomial T5(x), expanded at a=0, for
f(x) =
∫ x
0
ln |sect + tan t|dt.
Ans. T5(x) =
x2
2
+ x
4
4!
.
Problem 14
Suppose we only consider |x| ≤ 0.8. Find the best upper bound or maximum value
you can for∣∣∣sin x − (x − x33! + x55! )∣∣∣
Same question: If
(
x − x
3
3!
+ x
5
5!
)
is used to approximate sin x for |x| ≤ 0.8. What is
the maximum error? Explain what method you are using.
Problem 15
The Taylor polynomial T5(x) of degree 5 for (4 + x)
3/2 is
(4 + x)3/2 ≈ 8 + 3x + 3
16
x2 − 1
128
x3 + 3
4096
x4 − 3
32768
x5.
a) Use this polynomial to find Taylor polynomials for (4 + ...
SA Y SPEAK T ALK TELL
-Speak thường dùng khi 1 người nói với 1 tập thể
-Talk thường dùng khi 2 hay nhiều người đối thoại với nhau
-Say theo sau bởi words (cấu trúc: say something to somebody)
-Tell thường dùng để truyền tải thông tin (cấu trúc: tell somebody something)
Cụ thể:
. SAY:
Là động từ có tân ngữ, có nghĩa là”nói ra, nói rằng”, chú trọng nội dung được nói ra. Ex:
- Please say it again in English. (Làm ơn nói lại bằng tiếng Anh).
- They say that he is very ill. (Họ nói rằng cậu ấy ốm nặng).
. SPEAK:
Có nghĩa là “nói ra lời, phát biểu”, chú trọng mở miệng, nói ra lời.
Thường dùng làm động từ không có tân ngữ và cũng thường sử dụng với một giới từ ‘to‘, ‘about‘ hoặc ‘of‘ trước một tân ngữ. Khi có tân ngữ thì chỉ là một số ít từ chỉ thứ tiếng”truth” (sự thật).
Ex:
- He is going to speak at the meeting. (Anh ấy sẽ phát biểu trong cuộc mít tinh).
- I speak Chinese. I don’t speak Japanese. (Tôi nói tiếng Trung Quốc. Tôi không nói tiếng Nhật Bản).
- She spoke about her work at the university.
Bà ta nói về thành tích của mình tại trường Đại học.
- He spoke of his interest in photography.
Anh ta nói về sở thích về nhiếp ảnh.
Khi muốn “nói với ai” thì dùng speak to sb hay speak with sb.
Ex:
- She is speaking to our teacher. (Cô ấy đang nói chuyện với thày giáo của chúng ta).
. TELL:
Có nghĩa “kể, chú trọng, sự trình bày”. Thường gặp trong các kết cấu : tell sb sth (nói với ai điều gì đó), tell sb to do sth (bảo ai làm gì), tell sb about sth (cho ai biết về điều gì). Ex:
- The teacher is telling the class an interesting story. (Thầy giáo đang kể cho lớp nghe một câu chuyện thú vị).
- Please tell him to come to the blackboard. (Làm ơn bảo cậu ấy lên bảng đen).
- We tell him about the bad news. (Chúng tôi nói cho anh ta nghe về tin xấu đó).
. TALK:
Có nghĩa là”trao đổi, chuyện trò”, có nghĩa gần như speak, chú trọng động tác “nói’. Thường gặp trong các kết cấu: talk to sb (nói chuyện với ai), talk about sth (nói về điều gì), talk with sb (chuyện trò với ai).
Ex:
- What are they talking about? (Họ đang nói về chuyện gì thế?).
- He and his classmates often talk to eachother in English. (Cậu ấy và các bạn cùng lớp thường nói chuyện với nhau bằng tiếng Anh).”
Một Số Mẫu Câu Tiếng Anh Giao Tiếp Thông Dụng. 1. Right on! (Great!) - Quá đúng!
2. I did it! (I made it!) - Tôi thành công rồi!
3. Got a minute? - Có rảnh không?
4. About when? - Vào khoảng thời gian nào?
5. I won't take but a minute. - Sẽ không mất nhiều thời gian đâu.
6. Speak up! - Hãy nói lớn lên.
7. Seen Melissa? - Có thấy Melissa không?
8. So we've met again, eh? - Thế là ta lại gặp nhau phải không?
9. Come here. - Đến đây.
10. Come over. - Ghé chơi.
11. Don't go yet. - Đừng đi vội.
12. Please go first. After you. - Xin nhường đi trước. Tôi xin đi sau. 13. Thanks for letting me go first. - Cám ơn đã nhường đường.
14. What a relief. - Thật là nhẹ nhõm.
15. What the hell are you doing? - Anh đang làm cái quái gì thế kia?
13 quy tắc trọng âm trong tiếng anhdinhmyhuyenvi
Một Số Mẫu Câu Tiếng Anh Giao Tiếp Thông Dụng. 1. Right on! (Great!) - Quá đúng!
2. I did it! (I made it!) - Tôi thành công rồi!
3. Got a minute? - Có rảnh không?
4. About when? - Vào khoảng thời gian nào?
5. I won't take but a minute. - Sẽ không mất nhiều thời gian đâu.
6. Speak up! - Hãy nói lớn lên.
7. Seen Melissa? - Có thấy Melissa không?
8. So we've met again, eh? - Thế là ta lại gặp nhau phải không?
9. Come here. - Đến đây.
10. Come over. - Ghé chơi.
11. Don't go yet. - Đừng đi vội.
12. Please go first. After you. - Xin nhường đi trước. Tôi xin đi sau. 13. Thanks for letting me go first. - Cám ơn đã nhường đường.
14. What a relief. - Thật là nhẹ nhõm.
15. What the hell are you doing? - Anh đang làm cái quái gì thế kia?
Key Trends Shaping the Future of Infrastructure.pdfCheryl Hung
Keynote at DIGIT West Expo, Glasgow on 29 May 2024.
Cheryl Hung, ochery.com
Sr Director, Infrastructure Ecosystem, Arm.
The key trends across hardware, cloud and open-source; exploring how these areas are likely to mature and develop over the short and long-term, and then considering how organisations can position themselves to adapt and thrive.
GraphRAG is All You need? LLM & Knowledge GraphGuy Korland
Guy Korland, CEO and Co-founder of FalkorDB, will review two articles on the integration of language models with knowledge graphs.
1. Unifying Large Language Models and Knowledge Graphs: A Roadmap.
https://arxiv.org/abs/2306.08302
2. Microsoft Research's GraphRAG paper and a review paper on various uses of knowledge graphs:
https://www.microsoft.com/en-us/research/blog/graphrag-unlocking-llm-discovery-on-narrative-private-data/
Builder.ai Founder Sachin Dev Duggal's Strategic Approach to Create an Innova...Ramesh Iyer
In today's fast-changing business world, Companies that adapt and embrace new ideas often need help to keep up with the competition. However, fostering a culture of innovation takes much work. It takes vision, leadership and willingness to take risks in the right proportion. Sachin Dev Duggal, co-founder of Builder.ai, has perfected the art of this balance, creating a company culture where creativity and growth are nurtured at each stage.
Neuro-symbolic is not enough, we need neuro-*semantic*Frank van Harmelen
Neuro-symbolic (NeSy) AI is on the rise. However, simply machine learning on just any symbolic structure is not sufficient to really harvest the gains of NeSy. These will only be gained when the symbolic structures have an actual semantics. I give an operational definition of semantics as “predictable inference”.
All of this illustrated with link prediction over knowledge graphs, but the argument is general.
Software Delivery At the Speed of AI: Inflectra Invests In AI-Powered QualityInflectra
In this insightful webinar, Inflectra explores how artificial intelligence (AI) is transforming software development and testing. Discover how AI-powered tools are revolutionizing every stage of the software development lifecycle (SDLC), from design and prototyping to testing, deployment, and monitoring.
Learn about:
• The Future of Testing: How AI is shifting testing towards verification, analysis, and higher-level skills, while reducing repetitive tasks.
• Test Automation: How AI-powered test case generation, optimization, and self-healing tests are making testing more efficient and effective.
• Visual Testing: Explore the emerging capabilities of AI in visual testing and how it's set to revolutionize UI verification.
• Inflectra's AI Solutions: See demonstrations of Inflectra's cutting-edge AI tools like the ChatGPT plugin and Azure Open AI platform, designed to streamline your testing process.
Whether you're a developer, tester, or QA professional, this webinar will give you valuable insights into how AI is shaping the future of software delivery.
Connector Corner: Automate dynamic content and events by pushing a buttonDianaGray10
Here is something new! In our next Connector Corner webinar, we will demonstrate how you can use a single workflow to:
Create a campaign using Mailchimp with merge tags/fields
Send an interactive Slack channel message (using buttons)
Have the message received by managers and peers along with a test email for review
But there’s more:
In a second workflow supporting the same use case, you’ll see:
Your campaign sent to target colleagues for approval
If the “Approve” button is clicked, a Jira/Zendesk ticket is created for the marketing design team
But—if the “Reject” button is pushed, colleagues will be alerted via Slack message
Join us to learn more about this new, human-in-the-loop capability, brought to you by Integration Service connectors.
And...
Speakers:
Akshay Agnihotri, Product Manager
Charlie Greenberg, Host
Slack (or Teams) Automation for Bonterra Impact Management (fka Social Soluti...Jeffrey Haguewood
Sidekick Solutions uses Bonterra Impact Management (fka Social Solutions Apricot) and automation solutions to integrate data for business workflows.
We believe integration and automation are essential to user experience and the promise of efficient work through technology. Automation is the critical ingredient to realizing that full vision. We develop integration products and services for Bonterra Case Management software to support the deployment of automations for a variety of use cases.
This video focuses on the notifications, alerts, and approval requests using Slack for Bonterra Impact Management. The solutions covered in this webinar can also be deployed for Microsoft Teams.
Interested in deploying notification automations for Bonterra Impact Management? Contact us at sales@sidekicksolutionsllc.com to discuss next steps.
Accelerate your Kubernetes clusters with Varnish CachingThijs Feryn
A presentation about the usage and availability of Varnish on Kubernetes. This talk explores the capabilities of Varnish caching and shows how to use the Varnish Helm chart to deploy it to Kubernetes.
This presentation was delivered at K8SUG Singapore. See https://feryn.eu/presentations/accelerate-your-kubernetes-clusters-with-varnish-caching-k8sug-singapore-28-2024 for more details.
Essentials of Automations: Optimizing FME Workflows with ParametersSafe Software
Are you looking to streamline your workflows and boost your projects’ efficiency? Do you find yourself searching for ways to add flexibility and control over your FME workflows? If so, you’re in the right place.
Join us for an insightful dive into the world of FME parameters, a critical element in optimizing workflow efficiency. This webinar marks the beginning of our three-part “Essentials of Automation” series. This first webinar is designed to equip you with the knowledge and skills to utilize parameters effectively: enhancing the flexibility, maintainability, and user control of your FME projects.
Here’s what you’ll gain:
- Essentials of FME Parameters: Understand the pivotal role of parameters, including Reader/Writer, Transformer, User, and FME Flow categories. Discover how they are the key to unlocking automation and optimization within your workflows.
- Practical Applications in FME Form: Delve into key user parameter types including choice, connections, and file URLs. Allow users to control how a workflow runs, making your workflows more reusable. Learn to import values and deliver the best user experience for your workflows while enhancing accuracy.
- Optimization Strategies in FME Flow: Explore the creation and strategic deployment of parameters in FME Flow, including the use of deployment and geometry parameters, to maximize workflow efficiency.
- Pro Tips for Success: Gain insights on parameterizing connections and leveraging new features like Conditional Visibility for clarity and simplicity.
We’ll wrap up with a glimpse into future webinars, followed by a Q&A session to address your specific questions surrounding this topic.
Don’t miss this opportunity to elevate your FME expertise and drive your projects to new heights of efficiency.
Let's dive deeper into the world of ODC! Ricardo Alves (OutSystems) will join us to tell all about the new Data Fabric. After that, Sezen de Bruijn (OutSystems) will get into the details on how to best design a sturdy architecture within ODC.
Search and Society: Reimagining Information Access for Radical FuturesBhaskar Mitra
The field of Information retrieval (IR) is currently undergoing a transformative shift, at least partly due to the emerging applications of generative AI to information access. In this talk, we will deliberate on the sociotechnical implications of generative AI for information access. We will argue that there is both a critical necessity and an exciting opportunity for the IR community to re-center our research agendas on societal needs while dismantling the artificial separation between the work on fairness, accountability, transparency, and ethics in IR and the rest of IR research. Instead of adopting a reactionary strategy of trying to mitigate potential social harms from emerging technologies, the community should aim to proactively set the research agenda for the kinds of systems we should build inspired by diverse explicitly stated sociotechnical imaginaries. The sociotechnical imaginaries that underpin the design and development of information access technologies needs to be explicitly articulated, and we need to develop theories of change in context of these diverse perspectives. Our guiding future imaginaries must be informed by other academic fields, such as democratic theory and critical theory, and should be co-developed with social science scholars, legal scholars, civil rights and social justice activists, and artists, among others.
Search and Society: Reimagining Information Access for Radical Futures
Cc10 3 geo_har
1. 10.3 Geometric and Harmonic Series Contemporary Calculus 1
10.3 GEOMETRIC AND HARMONIC SERIES
This section uses ideas from Section 10.2 about series and their convergence to investigate some special
types of series. Geometric series are very important and appear in a variety of applications. Much of the early
th
work in the 17 century with series focused on geometric series and generalized them. Many of the ideas used
later in this chapter originated with geometric series. It is easy to determine whether a geometric series converges
or diverges, and when one does converge, we can easily find its sum. The harmonic series is important as an
example of a divergent series whose terms approach zero. A final type of series, called "telescoping," is discussed
briefly. Telescoping series are relatively uncommon, but their partial sums exhibit a particularly nice pattern.
∞
Geometric Series: ∑ C.r k = C + C.r + C.r2 + C.r3 + . . .
k=0
Example 1: Bouncing Ball: A "super ball" is thrown 10 feet
straight up into the air. On each bounce, it rebounds to
four fifths of its previous height (Fig. 1) so the sequence
of heights is 10 feet, 8 feet, 32/5 feet, 128/25 feet, etc.
(a) How far does the ball travel (up and down) during
th
its n bounce? (b) Use a sum to represent the total distance
traveled by the ball.
Solution: Since the ball travels up and down on each bounce,
the distance traveled during each bounce is twice the height
of the ball on that bounce so d1 = 2(10 feet) = 20 feet,
4
d2 = 16 feet, d3 = 64/5 feet, and, in general, dn = 5 .dn–1 . Looking at these values in another way,
4 4 4 4 4 2 4 4 4 2 4 3
d1 = 20 , d2 = 5 .(20) , d3 = 5 d2 = 5 .5 .20 = ( 5 ) (20) , d4 = 5 .d3 = 5 .( ( 5 ) .(20) ) = ( 5 ) .(20) ,
4 n–1 .
and, in general, dn = ( 5 ) (20) .
In theory, the ball bounces up and down forever, and the total distance traveled by the ball is the sum of the
distances traveled during each bounce (an up and down flight):
(first bounce) + (second bounce) + (third bounce) + (forth bounce) + . . .
4 4 2 4 3
= 20 + 5 (20) + ( 5 ) (20) + ( 5 ) (20) + . . .
∞
4 4 2 4 3 4 k
= 20.( 1 + 5 + ( 5 ) + ( 5 ) + . . . ) = 20. ∑ ( 5 ) .
k=0
2. 10.3 Geometric and Harmonic Series Contemporary Calculus 2
Practice 1: Cake: Three calculus students want to share a small square cake equally, but they go about it in a
rather strange way. First they cut the cake into 4 equal square pieces, each person takes one square, and one
square is left (Fig. 2). Then they cut the leftover piece into 4 equal square pieces, each person takes one square
and one square is left. And they keep repeating this process. (a) What fraction of the total cake does each
person "eventually" get? (b) Represent the amount of cake each person gets as a geometric series: (amount of
first piece) + (amount of second piece) + . . .
Each series in the previous Example and Practice problems is a Geometric series, a series in which each term is a fixed
multiple of the previous term. Geometric series have the form
∞ ∞
∑ C.r k = C + C.r + C.r2 + C.r3 + . . . = C. ∑ rk
k=0 k=0
with C ≠ 0 and r ≠ 0 representing fixed numbers. Each term in the series is r times the previous term. Geometric
series are among the most common and easiest series we will encounter. A simple test determines whether a
geometric series converges, and we can even determine the "sum" of the geometric series.
Geometric Series Theorem
∞ converges to 1
if | r | < 1
1–r
The geometric series ∑ rk 2 3
= 1 + r + r + r + ...
diverges if | r | ≥ 1
k=0
k k
Proof: If | r | ≥ 1, then | r | approaches 1 or +∞ as k becomes arbitrarily large, so the terms ak = r of
th
the geometric series do not approach 0. Therefore, by the n term test for divergence, the series diverges.
k
If | r | < 1, then the terms ak = r of the geometric series approach 0 so the series may or may not converge, and we
2 3 n
need to examine the limit of the partial sums sn = 1 + r + r + r + . . . + r of the series. For a geometric series, a
clever insight allows us to calculate those partial sums:
3. 10.3 Geometric and Harmonic Series Contemporary Calculus 3
(1 – r).sn = (1 – r).(1 + r + r + r + . . . + r )
2 3 n
= 1.(1 + r + r + r + . . . + r ) – r.(1 + r + r + r + . . . + r )
2 3 n 2 3 n
2 3 n 2 3 4 n n+1
= (1 + r + r + r + . . . + r ) – ( r + r + r + r + . . . + r + r )
n+1
= 1–r .
Since | r | < 1 we know r ≠ 1 so we can divide the previous result by 1 – r to get
n+1 n+1
2 3 n 1–r 1 r
sn = 1 + r + r + r + . . . + r = 1–r = 1–r – 1–r .
This formula for the nth partial sum of a geometric series is sometimes useful, but now we are interested in the
n+1
limit of sn as n approaches infinity. Since | r | < 1, r approaches 0 as n approaches infinity, so we can
n+1
1 r 1
conclude that the partial sums sn = 1 – r – 1 – r approach (as "n # $") .
1" r
∞
The geometric series ∑ r k converges to the value 1 when –1 < r < 1 .
1–r
k=0 !
∞ ∞ ∞
Finally, ∑ C.r k
= C. ∑ r
k
so we can easily determine whether or not ∑ C.r k converges and to
k=0 k=0 k=0
what number.
Example 2: How far did the ball in Example 1 travel?
4 4 2 4 3
Solution: The distance traveled, 20( 1 + 5 + ( 5 ) + ( 5 ) + . . . ) , is a geometric series with C = 20 and r =
4 4 2 4 3
4/5. Since | r | < 1, the series 1 + 5 + ( 5 ) + ( 5 ) + . . . converges to
1 1
1 – r = 1 – 4/5 = 5, so the total distance traveled is
4 4 2 4 3
20( 1 + 5 + ( 5 ) + ( 5 ) + . . . ) = 20( 5 ) = 100 feet.
Repeating decimal numbers are really geometric series in disguise, and we can use the Geometric Series Theorem
to represent the exact value of the sum as a fraction.
– ––
Example 3: Represent the repeating decimals 0.4 and 0.13 as geometric series and find their sums.
– 4 4 4 4 . ( 1 + 1 + ( 1 )2 + ( 1 )3 + . . . )
Solution: 0.4 = 0.444 . . . = 10 + 100 + 1000 + . . . = 10 10 10 10
which is a geometric series with a = 4/10 and r = 1/10. Since | r | < 1, the geometric series
4. 10.3 Geometric and Harmonic Series Contemporary Calculus 4
1 1 10 – 4 10 4
converges to 1–r = 1 – 1/10 = 9 , and 0.4 = 10 ( 9 ) = 9 .
–– 13 13 13
Similarly, 0.13 = 0.131313 ... = 100 + 10000 + 1000000 + . . .
13 1 1 1
.( 1 + 100 + ( 100 )2 + ( 100 )3 + . . . )
= 100
13 1 13 100 13
= 100 .( 1 – 1/100 ) = 100 ( 99 ) = 99 .
– –––
Practice 2: Represent the repeating decimals 0.3 and 0.432 as geometric series and find their sums.
One reason geometric series are important for us is that some series involving powers of x are geometric series.
∞ ∞
∑ k 2 ∑ k 2
Example 4: 3x = 3 + 3x + 3x + . . . and (2x – 5) = 1 + (2x – 5) + (2x – 5) + . . .
k=0 k=0
are geometric series with r = x and r = 2x – 5, respectively. Find the values of x for each series
so that the series converges.
Solution: A geometric series converges if and only if | r | < 1, so the first series converges if and
3
only if | x | < 1, or, equivalently, –1 < x < 1. The sum of the first series is 1 – x .
In the second series r = 2x – 5 so the series converges if and only if | 2x – 5 | < 1. Removing the absolute
value and solving for x, we get –1 < 2x – 5 < 1, and (adding 5 to each side and then dividing by 2) 2 < x < 3.
1
The second series converges if and only if 2 < x < 3. The sum of the second series is 1 – (2x – 5) or
1
6 – 2x .
∞ ∞
∑ k ∑ k
Practice 3: The series (2x ) and (3x – 4) are geometric series. Find the ratio r
k=0 k=0
for each series, and find all values of x for each series so that the series converges.
The series in the previous Example and Practice are called "power series" because they involve powers of the
variable x. Later in this chapter we will investigate other power series which are not geometric series
2 3
(e.g., 1 + x + x /2 + x /3 + . . . ), and we will try to find values of x which guarantee that the series converge.
5. 10.3 Geometric and Harmonic Series Contemporary Calculus 5
∞
1
Harmonic Series: ∑ k
k=1
∞
1
The series ∑ k is one of the best known and most important divergent series. It is called the harmonic series
k=1
because of its ties to music (Fig. 3).
If we simply calculate partial sums of the harmonic series, it is not clear that the series diverges ––
the partial sums sn grow, but as n becomes large,
the values of sn grow very, very slowly. Fig. 4 shows
the values of n needed for the partial sums sn to finally n sn
exceed the integer values 4, 5, 6, 8, 10, and 15. To 31 4.0224519544
examine the divergence of the harmonic series, brain power 83 5.00206827268
227 6.00436670835
is much more effective than a lot of computing power. 1,674 8.00048557200
12,367 10.00004300827
1,835,421 15.00000378267
Fig. 4
6. 10.3 Geometric and Harmonic Series Contemporary Calculus 6
We can show that the harmonic series is divergent by showing that the terms of the harmonic series can be
grouped into an infinite number of disjoint "chunks" each of which has a sum larger than 1/2. The series
∞
∑ 1 is clearly divergent because the partial sums grow arbitrarily large: by adding enough of the terms
2
k=1
together we can make the partial sums, sn > n/2 , larger than any predetermined number. Then we can conclude
that the partial sums of the harmonic series also approach infinity so the harmonic series diverges.
∞
1 1 1 1
Theorem: The harmonic series ∑ k = 1 + 2 + 3 + 4 + . . . diverges.
k=1
Proof: (This proof is essentially due to Oresme in 1630, twelve years before Newton was born. In 1821
Cauchy included Oresme's proof in a "Course in Analysis" and it became known as Cauchy's argument.)
1 1 1
Let S represent the sum of the harmonic series, S = 1 + 2 + 3 + 4 + . . . , and group the terms of the
series as indicated by the parentheses:
1
S=1+ 2 + (1
3
1
+ 4 )+(1
5
1 1 1
+ 6 + 7 + 8 )+(1
9
1 1
+ 10 + . . . + 16
1
) + ( 17 1
+. . . + 32 ) +...
↑ ↑ ↑ ↑
2 terms, each greater 4 terms, each greater 8 terms, each greater 16 terms, each greater
than or equal to 1/4 than or equal to 1/8 than or equal to 1/16 than or equal to 1/32
Each group in parentheses has a sum greater than 1/2, so
1
S>1 + 2 + ( 1 ) + ( 1 ) + ( 1 ) + . . . and
2 2 2
the sequence of partial sums { sn } does not converge to a finite number. Therefore, the harmonic
series diverges.
The harmonic series is an example of a divergent series whose terms, ak = 1/k, approach 0. If the terms of a
series approach 0, the series may or may not converge –– we need to investigate further.
Telescoping Series
Sailors in the seventeenth and eighteenth centuries used telescopes (Fig. 5)
which could be extended for viewing and collapsed for storing. Telescoping
series get their name because they exhibit a similar "collapsing" property.
Telescoping series are rather uncommon. But they are easy to analyze, and
it can be useful to recognize them.
7. 10.3 Geometric and Harmonic Series Contemporary Calculus 7
)
#1 1 & .
Example 5: Determine a formula for the partial sum sn of the series *% k " k + 1(
$ '
k=1
Then find lim sn . (Suggestion: It is tempting to algebraically consolidate terms,
n"#
but the pattern is clearer in this case if you first write out all of the terms.)
!
1
Solution: s1 = a1 = 1 – 2 . In later values of sn , part of each term cancels part of the next term:
!
# 1& # 1 1 & 1
s2 = a1 + a2 = %1" ( + % " ( = 1"
$ 2 ' $ 2 3' 3
# 1 & # 1 1& # 1 1 & 1
s3 = a1 + a2 + a3 = %1" ( + % " ( + % " ( = 1"
$ 2' $ 2 3' $ 3 4 ' 4
! In general, many of the pieces in each partial sum "collapse" and we are left with a simple form of sn :
!
# 1 & # 1 1& # 1 1& # 1 1 & 1
sn = a1 + a2 + ...+ an"1 + an = %1" ( + % " ( + ...+ % " (+% " ( = 1"
$ 2' $ 2 3' $ n - 1 n ' $ n n + 1' n +1
)
1 #1 1 &
Finally, lim sn = lim
n"# n"#
1 – n+1 = 1 so the series converges to 1: *% k " k + 1( = 1 .
$ '
! k=1
∞
Practice 4: Find the sum of the series ∑ [ sin( 1
k
1
) – sin( k + 1 ) ] .
! ! k=3 !
PROBLEMS
In problems 1 – 6, rewrite each geometric series using the sigma notation and calculate the value of the sum.
1 1 1 2 4 8
1. 1 + 3 + 9 + 27 + . . . 2. 1 + 3 + 9 + 27 + . . .
1 1 1 1 1 1 1 1
3. 8 + 16 + 32 + 64 + . . . 4. 1 – 2 + 4 – 8 + 16 – . . .
2 4 8 1 1 1
5. – 3 + 9 – 27 + . . . 6. 1+ e + 2 + 3 + ...
e e
k
7. Rewrite each series in the form of a sum of r , and then show that
1 1 1 1 1 1 1 1 1 1 1
(a) 2 + 4 + 8 + ... = 1, 3 + 9 + 27 + ... = 2 , and (b) for a > 1, a + 2 + 3 + ... = a – 1 .
a a
8. A ball is thrown 10 feet straight up into the air, and on each bounce, it rebounds to 60% of its
th
previous height. (a) How far does the ball travel (up and down) during its n bounce. (b) Use a sum to
represent the total distance traveled by the ball. (c) Find the total distance traveled by the ball.
8. 10.3 Geometric and Harmonic Series Contemporary Calculus 8
9. An old tennis ball is thrown 20 feet straight up into the air, and on each bounce, it rebounds to 40% of its
th
previous height. (a) How far does the ball travel (up and down) during its n bounce? (b) Use a sum to
represent the total distance traveled by the ball. (c) Find the total distance traveled by the ball.
10. Eighty people are going on an expedition by horseback through desolate country. The people and gear
require 90 horses, and additional horses are needed to carry food for the original 90 horses. Each additional
horse can carry enough food to feed 3 horses for the trip. How many additional horses are needed? (The
original 90 horses will require 30 extra horses to carry their food. The 30 extra horses require 10 more horses
to carry their food. etc.)
11. The mathematical diet you are following says you can eat "half of whatever is on the plate," so first you bite off
one half of the cake and put the other half back on the plate. Then you pick up the remaining half from the plate
(it’s "on the plate"), bite off half of that, and return the rest to the plate. And you continue this silly process of
picking up the piece from the plate, biting off half, and returning the rest to the plate. (a) Represent the total
amount you eat as a series. (b) How much of the cake is left after 1 bite, 2 bites, n bites? (c) "Eventually,"
how much of the cake do you eat?
12. Suppose in Fig. 6 we begin with a square with sides of length 1 (area = 1)
and construct another square inside by connecting the midpoints of the
sides. Then the new square has area 1/2. If we continue the process of
constructing each new square by connecting the midpoints of the sides of
the previous square, we get a sequence of squares each of which has 1/2
half the area of the previous square. Find the total area of all of the
squares.
13. Suppose in Fig. 7 we begin with a triangle with area 1 and construct
another triangle inside by connecting the midpoints of the sides.
Then the new triangle has area 1/4. Imagine that this construction
process is continued and find the total area
of all of the triangles.
14. Suppose in Fig. 8 we begin with a
circle of radius 1 and construct 2 more
circles inside, each with radius 1/2.
Continue the process of constructing two new circles inside each circle from the
previous step and find the total area of the circles.
9. 10.3 Geometric and Harmonic Series Contemporary Calculus 9
15. The construction of the Helga von Koch snowflake begins with an equilateral triangle of area 1 (Fig. 9).
Then each edge is subdivided into three equal lengths, and three equilateral triangles, each with area 1/9, are
built on these "middle thirds" adding a total of 3( 1/9 ) to the original area. The process is repeated: at the next
stage, 3.4 equilateral triangles, each with area 1/81, are built on the new "middle thirds" adding 3.4.(1/81)
more area. (a) Find the total area that results when this process is repeated forever.
number of edges # triangles added area of each triangle total area added
3 0 0 0
3.4 3 1/9 3/9
3.4 3.4 3.4/9 = (3/9)(4/9)
2 2 2
1/9
3.4 3.4 3.4 /9 = (3/9)(4 /9 )
3 2 3 2 3 2 2
1/9
3.4 3.4 3.4 /9 = (3/9)(4 /9 )
4 3 4 3 4 3 3
1/9
so the total added area of the snowflake is
2 3
3 3 4 3 4 3 4
1 + 9 + ( 9 )( 9 ) + ( 9 )( 2 ) + ( 9 )( 3 ) + . . .
9 9
(b) Express the perimeter of the Koch Snowflake as
a geometric series and find its sum.
(The area is finite, but the perimeter is infinite.)
16. Harmonic Tower: The base of a tower is a cube
whose edges are each one foot long. On top of it are cubes with edges of length
1/2, 1/3, 1/4, ... (Fig. 10).
(a) Represent the total height of the tower as a series. Is the height finite?
(b) Represent the total surface area of the cubes as a series.
(c) Represent the total volume of the cubes as a series.
(In the next section we will be able to determine if this surface area and
volume are finite or infinite.)
17. The base of a tower is a sphere whose radius is each one foot long. On top of
each sphere is another sphere with radius one half the radius of the sphere
immediately beneath it (Fig 11). (a) Represent the total height of the
tower as a series and find its sum. (b) Represent the total surface area
of the spheres as a series and find its sum. (c) Represent the total
volume of the spheres as a series and find its sum.
– ––
18. Represent the repeating decimals 0.6 and 0.63
as geometric series and find the value of each series as a simple fraction.
– – ––––––
19. Represent the repeating decimals 0.8 , 0.9 , and 0.285714 as geometric
series and find the value of each series as a simple fraction.
10. 10.3 Geometric and Harmonic Series Contemporary Calculus 10
– –– –––
20. Represent the repeating decimals 0.a , 0.a b , and 0.a b c as geometric series and find the value of
––––
each series as a simple fraction. What do you think the simple fraction representation is for 0.a b c d ?
In problems 21 – 32, find all values of x for which each geometric series converges.
∞ ∞ ∞
21. ∑ (2x + 1) k
22. ∑ (3 – x) k
23. ∑ (1 – 2x) k
k=1 k=1 k=1
∞ ∞ ∞
24. ∑ 5x
k
25. ∑ ( 7x ) k 26. ∑ ( x/3 ) k
k=1 k=1 k=1
2 3
x x x 2 4 8 2 3
27. 1 + 2 + 4 + 8 + ... 28. 1 + x + 2 + 3 + . . . 29. 1 + 2x + 4x + 8x + . . .
x x
∞ ∞ ∞ ∞
30. ∑ ( 2x/3 ) k
31. ∑ k
sin ( x ) 32. ∑ kx
e = ∑ ( ex ) k
k=1 k=1 k=1 k=1
2 3 1
33. One student thought the formula was 1 + x + x + x + . . . = 1 – x . The second student said "That
can't be right. If we replace x with 2, then the formula says the sum of the positive numbers
1
1 + 2 + 4 + 8 + ... is a negative number 1 – 2 = – 1." Who is right? Why?
34. The Classic Board Problem: If you have identical 1 foot long boards, they can be arranged to hang over
the edge of a table. One board can extend 1/2 foot beyond the edge (Fig. 12), two boards can extend
1/2 + 1/4 feet, and, in general, n boards can extend
1/2 + 1/4 + 1/6 + . . . + 1/(2n) feet beyond the edge.
(a) How many boards are needed
for an arrangement in which
the entire top board is beyond
the edge of the table?
(b) How many boards are needed
for an arrangement in which
the entire top two boards are
beyond the edge of the table?
(c) How far can an arrangement extend
beyond the edge of the table?
11. 10.3 Geometric and Harmonic Series Contemporary Calculus 11
In problems 35 – 40, calculate the value of the partial sum for n = 4 and n = 5 and find a formula for sn .
(The patterns may be more obvious if you do not simplify each term.)
) ) "
#1 1 & #1 1 &
35. * % k " k + 1(
$ '
36. * % k " k + 2(
$ '
37. # [k 3 $ (k + 1)3]
k=3 k=1 k=1
( " "
" k %
38. ) ln$ '
# k + 1&
39. # [f(k) $ f(k + 1)] 40. # [g(k) $ g(k + 2)]
k=1 k=3 k=1
! ! !
In problems 41 – 44, calculate s4 and s5 for each series and find the limit of sn as n approaches infinity.
! If the limit is a finite value, it represents the value of the infinite series. !
!
∞ ∞ ∞
1 1 1 1 1 1
41. ∑ sin( k ) – sin( k+1 ) 42. ∑ cos( k ) – cos( k+1 ) 43. ∑ 2 – 2
k=1 k=2 k=2 k (k+1)
∞ 1
1–k
44. ∑ ln( 1 – 12 ) 1
(Suggestion: Rewrite 1 – 2 as 1 )
k=3 k k 1 – k+1
Problems 45 and 46 are outlines of two "proofs by contradiction" that the harmonic series is divergent.
Each proof starts with the assumption that the "sum" of the harmonic series is a finite number, and then an
obviously false conclusion is derived from the assumption. Verify that each step follows from the assumption and
previous steps, and explain why the conclusion is false.
1 1 1 1 1 1
45. Assume that H = 1 + 2 + 3 + 4 + 5 + 6 + 7 + . . . is a finite number, and let
1 1 1 1 1 1 1
O = 1 + 3 + 5 + 7 + . . . be the sum of the "odd reciprocals," and E = 2 + 4 + 6 + 8 + . . .
be the sum of the "even reciprocals." Then
(i) H = O + E, (ii) each term of O is larger than the corresponding term of E so O > E,
1 1 1 1 1 1 1 1 1
and (iii) E= 2 + 4 + 6 + 8 +...= 2 {1+2 + 3 + 4 +...} =2 H.
1 1
Therefore H = O + E > 2 H + 2 H = H (so "H is strictly bigger than H," a contradiction).
1 1 1 1 1 1
II. Assume that H = 1 + 2 + 3 + 4 + 5 + 6 + 7 + . . . is a finite number, and, starting with the
1 1 1 1
second term, group the terms into groups of three. Then, using n–1 + n + n+1 > 3. n , we have
1 1 1 1 1 1 1 1 1
H = 1 + ( 2 + 3 + 4 ) + ( 5 + 6 + 7 ) + ( 8 + 9 + 10 ) + . . .
1 1
>1+ ( 1 ) + ( 2 ) + ( 3 ) +...
1 1 1
= 1 + ( 1 + 2 + 3 + 4 + . . . ) = 1 + H . Therefore, H > 1 + H (so "H is bigger than 1 + H").
12. 10.3 Geometric and Harmonic Series Contemporary Calculus 12
46. Jacob Bernoulli (1654–1705) was a master of understanding and manipulating series by breaking a difficult series
into a sum of easier series. He used that technique to find the sum of the non–geometric series
∞
k
∑ k
k
= 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + . . . + k/2 + . . . in his book Ars Conjectandi , 1713.
k=1 2
n
Show that 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + . . . + n/2 + . . . can be written as
n
1/2 + 1/4 + 1/8 + 1/16 + 1/32 + . . . + 1/2 + . . . (a)
n
plus 1/4+ 1/8 + 1/16 + 1/32 + . . . + 1/2 + . . . (b)
n
plus 1/8 + 1/16 + 1/32 + . . . + 1/2 + . . . (c)
plus . . . etc.
Find the values of the geometric series (a), (b), (c), etc. and then find the sum of these values (another geometric
series).
47. Bernoulli's approach in problem 46 can also be interpreted as a geometric argument for representing the area
in Fig. 13 in two different ways. (a) Represent the total area in Fig. 13a as a (geometric) sum of the areas of
the side–by–side rectangles, and find the sum of the series. (b) Represent the total area of the stacked
rectangles in Fig. 13b as a sum of the areas of the horizontal slices.
Since both series represent the same total area, the values of the series are equal.
13. 10.3 Geometric and Harmonic Series Contemporary Calculus 13
48. Use the approach of problem 46 to find a formula for the sum of
∞
(a) the value of ∑ kk = 1/3 + 2/9 + 3/27 + 4/81 + . . . + k/3k + . . . and
k=13
∞
k
(b) a formula for the value of ∑ k
2 3 4 k
= 1/c + 2/c + 3/c + 4/c + . . . + k/c + . . . for c > 1.
k=1 c
2
(answers: (a) 3/4, (b) c/(c–1) )
Practice Answers
Practice 1: (a) Since they each get equal shares, and the whole cake is distributed, they each get 1/3 of the cake.
2
More precisely, after step 1, 1/4 of the cake remains and 3/4 was shared. After step 2, (1/4) of the cake
2 n
remains and 1 – (1/4) was shared. After step n, (1/4) of the cake remains and
n 1 n
1 – (1/4) was shared. So after step n, each student has ( 3 )( 1 – (1/4) ) of the cake. "Eventually,"
1
each student gets (almost) 3 of the cake.
1 1 2 1 3 1 1 1 2
(b) ( 4 ) + ( 4 ) + ( 4 ) + ... = ( 4 ) { 1 + ( 4 ) + ( 4 ) + ... }
– 3 3 3 3 1 1 1
.( 1 + 10 + ( 10 )2 + ( 10 )3 + . . . )
Practice 2: 0.3 = 0.333 ... = 10 + 100 + 1000 + . . . = 10
which is a geometric series with a = 3/10 and r = 1/10. Since | r | < 1, the geometric series
1 1 10 – 3 10 3 1
converges to 1 – r = 1 – 1/10 = 9 , and 0.3 = 10 ( 9 ) = 9 = 3 .
––– 432 432 432
Similarly, 0.432 = 0.432432432 ... = 1000 + 1000000 + 1000000000 + . . .
432 1 1 1
.( 1 + 1000 + ( 1000 )2 + ( 100 )3 + . . . )
= 1000
432 1 432 1000 432 16
= 1000 .( 1 – 1/1000 ) = 1000 ( 999 ) = 999 = 37 .
Practice 3: r = 2x : If |2x| < 1, then –1 < 2x < 1 so –1/2 < x < 1/2.
∞
∑ (2x ) k converges (to 1 –12x ) when –1/2 < x < 1/2.
k=0
r = 3x – 4 : If |3x – 4| < 1, then –1 < 3x – 4 < 1 so 3 < 3x < 5 and 1 < x < 5/3.
∞
∑ (3x – 4) k converges (to 1 – (3x – 4) = 5 –13x ) when 1 < x < 5/3 .
1
k=0
14. 10.3 Geometric and Harmonic Series Contemporary Calculus 14
n
1 1
Practice 4: Let sn = ∑ sin( k ) – sin( k + 1 )
k=3
1 1 1 1 1 1 1 1
= {sin( 3 ) – sin( 4 ) } + {sin( 4 ) – sin( 5 ) } + {sin( 5 ) – sin( 6 ) } + ... + {sin( n ) – sin( n+1 ) }
1 1
= sin( 3 ) – sin( n+1 ) .
1 1 1 1
Then lim sn = lim {sin( 3 ) – sin( n+1 )} = sin( 3 ) so the series converges to sin( 3 ):
n"# n"#
∞
1 1 1
∑ sin( k ) – sin( k + 1 ) = sin( 3 ) ≈ 0.327 .
k=1
! !
Appendix: MAPLE and WolframAlpha for Partial Sums of Geometric Series
MAPLE command for
100
3
" n
: sum(3*(1/2)^n , n=0..100) ; (then press ENTER key)
n=0 2
"
3
# n
: sum(3*(1/2)^n , n=0..infinity) ; (then press ENTER key)
n=0 2
!
WolframAlpha (free at http://www.wolframalpha.com )
! Asking sum 3/2^k , k = 1 to infinity
gives the sum of the infinite series and a graph of the partial sums.
Asking sum 3/2^k , k = 1 to 100
gives the sum as an exact fraction (strange) and as a decimal.