1. Devious Dingo
Bleached Bones, Australia
11 September 2015
Math 225 Volan Nnanpalle
Bloomsburg University
Bloomsburg, PA 17815
Dear Devious Dingo:
I received your letter about your recurring nightmares. No one
should have to go through sleepless nights. You were sent to the right
student. I will clearly answer all your questions and state proofs for the
test being used in a well-defined and concise manner.
These are your questions I plan to answer in my report:
“Do I ever get to Emu?”
Mr. Dingo you will get Emu.
“Do I have any chance?”
Unfortunately, you don’t have a chance due to how long it takes you to
get Emu and the restrictions of the human body.
“If I do get there, how long does it take?”
It will take you about 1.97007114 * 10434
seconds or
6.247054522 * 10426
years
“Should I take a snack to eat along the way?”
If you’re willing to make this quest to get Emu, the you will have to
pack snacks that will last you about 7 years.
Now proceed to the next page to receive piece of mine and finally
be able to get a goodnight’s sleep.

2. Harmonic Test Proof
Let x ∈ ℝ
Let P(x) = 𝐻2 𝑥 ≥ 1+
𝑥
2
Let x = 0. So P(0) is
𝐻20 = 𝐻1 = 1 ≥ 1 +
0
2
Assume that P(x) is true, so that
= 𝐻2 𝑥 ≥ 1 +
𝑥
2
.
It must be shown that P(x + 1), which states
𝐻2 𝑥+1 ≥ 1 +
𝑥+1
2
,
must also be true under this assumption. This is done as follows:
𝐻2 𝑥+1 = 1 +
1
2
+
1
3
+. . . +
1
2 𝑥 +
1
2 𝑥+1
+ . . . +
1
2 𝑥+1
= 1 +
1
2
+
1
3
+. . . +
1
2 𝑥 +
1
2 𝑥+1
+ . . . +
1
2 𝑥+1
𝐻2 𝑥
= 𝐻2 𝑥 +
1
2 𝑥+1
+ . . . +
1
2 𝑥+1
≥ (1 +
𝑥
2
) +
1
2 𝑥+1
+ . . . +
1
2 𝑥+1

3. ≥ (1 +
𝑥
2
) +
1
2 𝑥 +
1
2 𝑥+1
≥ (1 +
𝑥
2
) +
1
2
= 1 +
𝑥+1
2
Using the Principal of Induction, this proof shows that the inequality
for the harmonic series is valid for all nonnegative integers x.
Source: Discrete Mathematics
Integral Test
Let n ∈ ℕ
Source: Integral Test Proof

4. Let n ∈ ℕ
n = seconds
𝒅 𝒏= distance between Devious & Emu after n seconds
Solution
“Do I ever get Emu?”
𝑑 𝑛+1 = 𝑑 𝑛 + (% of distance ahead) * 1000 -1
= 𝑑 𝑛 +
𝑑 𝑛
𝑛∗1000
* 1000 -1
= 𝑑 𝑛+
𝑑 𝑛
𝑛
- 1
= (n +1) * (
𝑑 𝑛
𝑛
-
1
𝑛+1
)
𝑑0 = 1998
𝑑1 = 1(𝑑0 − 1) = 1997
𝑑2 = 2(𝑑1 −
1
2
) = 3993
𝑑3 = 3(
𝑑2
2
−
1
3
) = 5988.5
𝑑 𝑛= n (𝑑0 – ∑
1
𝑘
𝑛
𝑘=1 ) , Harmonic Series
The Harmonic series diverges, so we will eventually get ∑
1
𝑘
∞
𝑘=1 ≥ 𝑑0
where 𝑑 𝑛 ≤ 0 and you would catch emu. YOU WILL GET EMU!!

5. “If I do get there, how long does it take?”
Integral Test
∑
1
𝑘
𝑛+1
𝑘=1 > ∫
1
𝑥
𝑑 𝑥
𝑛+1
1
Solution
∫
1
𝑥
𝑑 𝑥
𝑛+1
1
> 1000 , gives n = 𝑒1000
− 1 ≈ 1.97007114 * 10434
seconds
or 6.247054522 * 10426
years
To show the general case,
Let t = distance traveled per second,
𝑑 𝑛 = 𝑛(𝑑0 − 𝑡 ∑
1
𝑘
𝑛
𝑘=1 ),
You will get Emu in 𝑒
𝑑0
𝑡 − 1 seconds.

6. Work Cited
"Integral Test Proof." Calculus. N.p., n.d. Web. 16 Sept. 2015.
1. MATH 2420 Discrete Mathematics (n.d.): n. pag. Web.