The document provides information on various integration techniques including the midpoint rule, trapezoidal rule, Simpson's rule, integration by parts, trigonometric substitutions, and applications of integrals such as finding the area between curves, arc length, surface area of revolution, and volume of revolution. It also covers integrals of common functions, properties of integrals, and techniques for parametric and polar coordinates.

2. Integrals
Midpoint rule
∫
b
a
f(x) dx ≈
n
∑
i=1
f (x̄i) Δx = Δx [f ( ¯
x1) + … + f ( ¯
xn)]
where (xi−1, xi) and
Δx =
b − a
n
and
x̄i =
1
2
(xi−1 + xi)
Trapezoidal rule
∫
b
a
f(x) dx ≈ Tn =
Δx
2
[f(x0) + 2f(x1) + 2f(x2) + … + 2f(xn−1) + f(xn)]
where Δx =
b − a
n
and
xi = a + iΔx
Midpoint and trapezoidal error bounds
ET and EM are the errors in the trapezoidal and midpoint rules
1

3. |ET | ≤
K(b − a)3
12n2
and |EM | ≤
K(b − a)3
24n2
where | f′′(x)| ≤ K
for a ≤ x ≤ b
Simpson’s rule
∫
b
a
f(x) dx ≈ Sn =
Δx
3
[f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + … + 2f(xn−2) + 4f(xn−1) + f(xn)]
where n is even and
Δx =
b − a
n
Simpson’s error bounds
ES is the error in Simpson’s rule
|ES | ≤
K(b − a)5
180n4
where f (4)
(x) ≤ K
for a ≤ x ≤ b
2

4. Symmetric functions
Suppose f is continuous on [−a, a].
If f is even [f(−x) = f(x)], then
∫
a
−a
f(x) dx = 2
∫
a
0
f(x) dx
If f is odd [f(−x) = − f(x)], then
∫
a
−a
f(x) dx = 0
Limit process for area under the curve
∫
b
a
f(x) dx = lim
n→∞
n
∑
i=1
f(xi)Δx
where Δx =
b − a
n
and xi = a + iΔx
Summation formulas for the limit process
n
∑
i=1
k = kn where k is any non-zero constant
n
∑
i=1
i =
n(n + 1)
2
=
n2
+ n
2
3

5. n
∑
i=1
i2
=
n(n + 1)(2n + 1)
6
=
2n3
+ 3n2
+ n
6
n
∑
i=1
i3
=
n2
(n + 1)2
4
=
n4
+ 2n3
+ n2
4
n
∑
i=1
i4
=
n(2n + 1)(n + 1)(3n2
+ 3n − 1)
30
=
6n5
+ 15n4
+ 10n3
+ 6n2
− n
30
Net change theorem
∫
b
a
F′(x) dx = F(b) − F(a)
Fundamental theorem of calculus
Suppose f is continuous on [a, b].
Part 1
Given integral How to solve it
f(x) =
∫
x
a
f(t) dt Plug x in for t.
f(x) =
∫
a
x
f(t) dt Reverse limits of integration and multiply by
−1, then plug x in for t.
4

6. f(x) =
∫
g(x)
a
f(t) dt Plug g(x) in for t, then multiply by dg/dx.
f(x) =
∫
a
g(x)
f(t) dt Reverse limits of integration and multiply by
−1, then plug g(x) in for t and multiply by dg/dx.
f(x) =
∫
h(x)
g(x)
f(t) dt Split the limits of integration as
∫
0
g(x)
f(t) dt +
∫
h(x)
0
f(t) dt. Reverse limits of
integration on
∫
0
g(x)
f(t) dt and multiply by −1,
then plug g(x) and h(x) in for t, multiplying by
dg/dx and dh/dx respectively.
Part 2
∫
b
a
f(x) dx = F(b) − F(a)
where F is any antiderivative of f, that is, a function such that
F′ = f
Integration by parts
∫
u dv = uv −
∫
v du
5

7. Properties of integrals
∫
b
a
c dx = c(b − a)
∫
b
a
f(x) + g(x) dx =
∫
b
a
f(x) dx +
∫
b
a
g(x) dx
∫
b
a
cf(x) dx = c
∫
b
a
f(x) dx
∫
b
a
f(x) − g(x) dx =
∫
b
a
f(x) dx −
∫
b
a
g(x) dx
Common indefinite integrals
∫
k dx = kx + C
∫
xn
dx =
xn+1
n + 1
+ C with n ≠ − 1
∫
1
x
dx = ln|x| + C
∫
ex
dx = ex
+ C
∫
ax
dx =
ax
ln a
+ C
6

8. Integrals of trig functions
∫
sin x dx = − cos x + C
∫
csc x dx = ln csc x − cot x + C
or
∫
csc x dx = ln
(
sin
x
2)
− ln
(
cos
x
2)
+ C
∫
cos x dx = sin x + C
∫
sec x dx = ln sec x + tan x + C
or
∫
sec x dx = ln
(
sin
x
2
+ cos
x
2)
− ln
(
cos
x
2
− sin
x
2)
+ C
∫
tan x dx = − ln cos x + C
∫
cot x dx = ln sin x + C
Other common trig integrals
∫
sec2
x dx = tan x + C
∫
csc2
x dx = − cot x + C
∫
sec x tan x dx = sec x + C
∫
csc x cot x dx = − csc x + C
∫
1
x2 + 1
dx = tan−1
x + C
∫
1
1 − x2
dx = sin−1
x + C
∫
sinh x dx = cosh x + C
∫
cosh x dx = sinh x + C
7

9. Integrals of inverse hyperbolic trig functions
∫
sinh−1
x dx = x sinh−1
x − x2
+ 1 + C
∫
cosh−1
x dx = x cosh−1
x − x − 1 x + 1 + C
∫
tanh−1
x dx =
1
2
log(1 − x2
) + x tanh−1
x + C
∫
coth−1
x dx =
1
2
log(1 − x2
) + x coth−1
x + C
Integrals resulting in inverse hyperbolic trig functions
∫
1
x2 + 1
dx = sinh−1
x
∫
1
x − 1 x + 1
dx = cosh−1
x
∫
1
1 − x2
dx = tanh−1
x
∫
1
1 − x2
dx = coth−1
x
8

10. Trig substitution setup
sin tan sec
the integral includes a2
− u2
a2
+ u2
u2
− a2
so substitute u = a sin θ u = a tan θ u = a sec θ
and use the identity 1 − sin2
θ = cos2
θ 1 + tan2
θ = sec2
θ sec2
θ − 1 = tan2
θ
solve for the trig sin θ =
u
a
tan θ =
u
a
sec θ =
u
a
and for du du = a cos θ dθ du = a sec2
θ dθ du = a sec θ tan θ dθ
and for θ θ = arcsin
u
a
θ = arctan
u
a
θ = arcsec
u
a
reference triangle
9
a
a
u
u u
a
a2
− u2
a2
+ u2
u2
− a2
θ θ θ

11. Trig substitution simplification
arcsin x arccos x arctan x arccscx arcsecx arccotx
sin of x 1 − x2 x
x2 + 1
1
x
1 −
1
x2
−
1
x
1
x2
+ 1
cos of 1 − x2
x
1
x2 + 1
1 −
1
x2
1
x
1
1
x2
+ 1
tan of
x
1 − x2
1 − x2
x
x
1
x 1 −
1
x2
x 1 −
1
x2
1
x
csc of
1
x
1
1 − x2
x2
+ 1
x
x
1
1 −
1
x2
x
1
x2
+ 1
sec of
1
1 − x2
1
x
x2
+ 1
1
1 −
1
x2
x
1
x2
+ 1
cot of
1 − x2
x
x
1 − x2
1
x
x 1 −
1
x2
1
x 1 −
1
x2
x
10

12. Applications of Integrals
Average value
fave =
1
b − a ∫
b
a
f(x) dx
Area between curves
A =
∫
b
a
| f(x) − g(x)| dx
Arc length
L =
∫
b
a
1 + [f′(x)]
2
dx =
∫
b
a
1 +
(
dy
dx)
2
dx for a function y = f(x)
on the interval a ≤ x ≤ b
L =
∫
d
c
1 + [g′(y)]
2
dy =
∫
d
c
1 +
(
dx
dy )
2
dy for a function x = g(y)
on the interval c ≤ y ≤ d
11

13. Surface area of revolution
The surface area of revolution is
S =
∫
b
a
2πy 1 +
(
dy
dx)
2
dx for a function y = f(x)
rotated about the x-axis
on the interval a ≤ x ≤ b
S =
∫
b
a
2πx 1 +
(
dy
dx )
2
dx for a function y = f(x)
rotated about the y-axis
on the interval a ≤ x ≤ b
S =
∫
d
c
2πy 1 +
(
dx
dy )
2
dy for a function x = g(y)
rotated about the x-axis
on the interval c ≤ y ≤ d
S =
∫
d
c
2πx 1 +
(
dx
dy )
2
dy for a function x = g(y)
rotated about the y-axis
on the interval c ≤ y ≤ d
12

14. Volume of revolution
Axis Disks Washers Shells
∫
area width
∫
area width
∫
circumference height width
Axis of revolution: HORIZONTAL
x-axis
∫
b
a
π [f(x)]
2
dx
∫
b
a
π [f(x)]
2
− π [g(x)]
2
dx
∫
d
c
2πy [f(y) − g(y)] dy
y = − k
∫
b
a
π [k + f(x)]
2
− π [k + g(x)]
2
dx
∫
d
c
2π(y + k)[f(y) − g(y)] dy
y = k
∫
b
a
π [k − g(x)]
2
− π [k − f(x)]
2
dx
∫
d
c
2π(k − y)[f(y) − g(y)] dy
Axis of revolution: VERTICAL
y-axis
∫
d
c
π [f(y)]
2
dy
∫
d
c
π [f(y)]
2
− π [g(y)]
2
dy
∫
b
a
2πx [f(x) − g(x)] dx
x = − k
∫
d
c
π [k + f(y)]
2
− π [k + g(y)]
2
dy
∫
b
a
2π(x + k)[f(x) − g(x)] dx
x = k
∫
d
c
π [k − g(y)]
2
− π [k − f(y)]
2
dy
∫
b
a
2π(k − x)[f(x) − g(x)] dx
13

15. Mean value theorem for integrals
If f is continuous on [a, b], then c exists in [a, b] such that
f(c) = fave =
1
b − a ∫
b
a
f(x) dx, that is,
∫
b
a
f(x) dx = f(c)(b − a)
Moments of the region
The moment of the region
about the y-axis is My = lim
n→∞
n
∑
i=1
ρx̄i f(x̄i)Δx = ρ
∫
b
a
xf(x) dx
about the x-axis is Mx = lim
n→∞
n
∑
i=1
ρ ⋅
1
2
[f(x̄i)]
2
Δx = ρ
∫
b
a
1
2
[f(x)]
2
dx
Center of mass of the region
The center of mass is located at (x̄, ȳ) where
x̄ =
My
m
=
1
A ∫
b
a
xf(x) dx and ȳ =
Mx
m
=
1
A ∫
b
a
1
2
[f(x)]
2
dx
14

16. Center of mass of the region bounded by two curves
The center of mass is located at (x̄, ȳ) where
x̄ =
1
A ∫
b
a
x [f(x) − g(x)] dx and ȳ =
1
A ∫
b
a
1
2 {[f(x)]
2
− [g(x)]
2
} dx
15

17. Polar & Parametric
Parametric
Derivatives
dy
dx
=
dy
dt
dx
dt
where
dx
dt
≠ 0
Area
A =
∫
b
a
y dx =
∫
β
α
g(t)f′(t) dt
Surface area of revolution
The surface area of a parametric curve rotated
about the x-axis is Sx =
∫
b
a
2πy
(
dx
dt )
2
+
(
dy
dt )
2
dt
about the y-axis is Sy =
∫
b
a
2πx
(
dx
dt )
2
+
(
dy
dt )
2
dt
16

18. Arc length
L =
∫
β
α
(
dx
dt )
2
+
(
dy
dt )
2
dt
Volume of revolution
The volume created by rotating a parametric curve
about the x-axis is Vx =
∫
b
a
πy2
[x′(t)] dt
about the y-axis is Vy =
∫
b
a
πx2
[y′(t)] dt
Polar
Conversion between cartesian and polar coordinates
x = r cos θ y = r sin θ x2
+ y2
= r2
Distance between two points
The distance between two polar coordinate points (r1, θ1) and (r2, θ2) is
17

19. D = (r2 cos θ2 − r1 cos θ1)
2
+ (r2 sin θ2 − r1 sin θ1)
2
Area
The area enclosed by a polar curve is
A =
∫
b
a
1
2
[f(θ)]
2
dθ =
∫
b
a
1
2
r2
dθ
Area between curves
The area between two polar curves is
A =
1
2 ∫
β
α
(router)2
− (rinner)2
dθ
Arc length
L =
∫
b
a
r2
+
(
dr
dθ )
2
dθ
Surface area of revolution
The surface area of revolution of a polar parametric curve rotated
18

20. about the x-axis is Sx =
∫
β
α
2πy r2
+
(
dr
dθ)
2
dθ
about the y-axis is Sy =
∫
β
α
2πx r2
+
(
dr
dθ)
2
dθ
19

21. Sequences & Series
Limit of a sequence
The limit of a sequence {an} is L
lim
n→∞
an = L or an → L as n → ∞
if we can make the terms of an closer and closer to L as we we make n
larger and larger. If
lim
n→∞
an
exists, the sequence converges (is convergent). Otherwise it diverges (is
divergent).
Precise definition of the limit of a sequence
The limit of a sequence {an} is L
lim
n→∞
an = L or an → L as n → ∞
if for every ϵ > 0 there is a corresponding integer N such that
if n > N then |an − L| < ϵ
20

22. Limit laws for sequences
If {an} and {bn} are convergent sequences and c is a constant, then
lim
n→∞
(an + bn) = lim
n→∞
an + lim
n→∞
bn lim
n→∞
(an − bn) = lim
n→∞
an − lim
n→∞
bn
lim
n→∞
can = c lim
n→∞
an lim
n→∞
c = c
lim
n→∞
(anbn) = lim
n→∞
an ⋅ lim
n→∞
bn lim
n→∞
an
bn
=
limn→∞ an
limn→∞ bn
if
lim
n→∞
bn ≠ 0
lim
n→∞
ap
n =
[
lim
n→∞
an]
p
if p > 0 and an > 0
Squeeze theorem for sequences
If an ≤ bn ≤ cn for n ≥ n0 and lim
n→∞
an = lim
n→∞
cn = L then lim
n→∞
bn = L.
Absolute value of a sequence
If lim
n→∞
|an | = 0, then lim
n→∞
an = 0.
21

23. Convergence of a sequence rn
The sequence {rn
} is convergent if −1 < r ≤ 1 and divergent for all other
values of r.
lim
n→∞
rn
= 0 if −1 < r < 1
= 1 if r = 1
Increasing, decreasing, and monotonic sequences
A sequence {an} is
increasing if an < an+1 for all n ≥ 1, (a1 < a2 < a3 < . . . )
decreasing if an > an+1 for all n ≥ 1, (a1 > a2 > a3 > . . . )
monotonic if it’s either increasing or decreasing
Bounded sequences
A sequence {an} is
bounded above if there’s a number M such that an ≤ M for all n ≥ 1
bounded below if there’s a number m such that m ≤ an for all n ≥ 1
a bounded sequence if it’s bounded above and below
22

24. Monotonic sequence theorem
Every bounded, monotonic sequence is convergent.
Partial sum of the series
Given a series
∞
∑
n=1
an = a1 + a2 + a3 + . . . , let sn denote its nth partial sum.
sn =
n
∑
i=1
ai = a1 + a2 + … + an
If the sequence {sn} converges and lim
n→∞
sn = s exists as a real number, then
the series
∑
an converges and we can say
a1 + a2 + … + an + … = s or
∞
∑
n=1
an = s
The number s is the sum of the series. If the sequence {sn} diverges, then
the series diverges.
Convergence and sum of a geometric series
The geometric series
∞
∑
n=1
arn−1
= a + ar + ar2
+ . . .
23

25. converges if |r| < 1, otherwise it diverges (if |r| ≥ 1). The sum of the
convergent series is
∞
∑
n=1
arn−1
=
a
1 − r
|r| < 1
Convergence of an
If the series
∞
∑
n=1
an converges, then lim
n→∞
an = 0.
Test for divergence
If lim
n→∞
an doesn’t exist or if lim
n→∞
an ≠ 0, then the series
∞
∑
n=1
an diverges.
Laws of convergent series
If the series an and bn both converge, then so do these (where c is a
constant):
∞
∑
n=1
can = c
∞
∑
n=1
an
∞
∑
n=1
(an + bn) =
∞
∑
n=1
an +
∞
∑
n=1
bn
24

26. ∞
∑
n=1
(an − bn) =
∞
∑
n=1
an −
∞
∑
n=1
bn
Integral test for convergence
Suppose f is a continuous, positive, decreasing function on [1,∞), and let
an = f(n).
If
∫
∞
1
f(x) dx converges, then
∞
∑
n=1
an converges.
If
∫
∞
1
f(x) dx diverges, then
∞
∑
n=1
an diverges.
Remainder estimate for the integral test
Suppose f(k) = ak, where f is a continuous, positive, decreasing function for
x ≥ n and
∑
an converges. If Rn = s − sn, then
∫
∞
n+1
f(x) dx ≤ Rn ≤
∫
∞
n
f(x) dx
p-Series test for convergence
The p-series
∞
∑
n=1
1
np
25

27. converges if p > 1
diverges if p ≤ 1
Comparison test for convergence
Suppose that
∑
an and
∑
bn are series with positive terms.
If
∑
bn converges and an ≤ bn for all n, then
∑
an converges.
If
∑
bn diverges and an ≥ bn for all n, then
∑
an diverges.
Limit comparison test for convergence
Suppose that
∑
an and
∑
bn are series with positive terms.
If lim
n→∞
an
bn
= c
where c is a finite number and
where 0 < c < ∞
then either both series converge or both diverge.
Alternating series test for convergence
If the alternating series
26

28. ∞
∑
n=1
(−1)n−1
bn = b1 − b2 + b3 − b4 + b5 − b6 + … bn > 0
satisfies
bn+1 ≤ bn for all n
lim
n→∞
bn = 0
then the series converges.
Alternating series estimation theorem
If s =
∑
(−1)n−1
bn is the sum of an alternating series that satisfies
bn+1 ≤ bn
lim
n→∞
bn = 0
then |Rn | = |s − sn | ≤ bn+1.
Absolute convergence
A series
∑
an is absolutely convergent if the series of absolute values
∑
|an | is convergent.
If a series
∑
an is absolutely convergent, then it’s convergent.
27

29. Conditional convergence
A series
∑
an is conditionally convergent if it’s convergent but not
absolutely convergent.
Ratio test for convergence
If lim
n→∞
an+1
an
= L < 1, then the series
∞
∑
n=1
an is absolutely convergent.
If lim
n→∞
an+1
an
= L > 1 or lim
n→∞
an+1
an
= ∞, then the series
∞
∑
n=1
an is divergent.
If lim
n→∞
an+1
an
= 1, the ratio test is inconclusive about the convergence of
∞
∑
n=1
an, which means we’ll have to use a different convergence test to
determine convergence.
Root test for convergence
If lim
n→∞
n
|an | = L < 1, then the series
∞
∑
n=1
an is absolutely convergent.
If lim
n→∞
n
|an | = L > 1 or lim
n→∞
n
|an | = ∞, then the series
∞
∑
n=1
an is divergent.
28

30. If lim
n→∞
n
|an | = 1, the root test is inconclusive about the convergence of
∞
∑
n=1
an, which means we’ll have to use a different convergence test to
determine convergence.
Convergence of power series
Given a power series
∞
∑
n=0
cn(x − a)n
,
the series converges only when x = a or
the series converges for all x or
there is a positive number R such that the series converges if |x − a| < R and
diverges if |x − a| > R
Differentiation and integration of power series
If the power series
∑
cn(x − a)n
has a radius of convergence R > 0, then the
function f defined by
f(x) = c0 + c1(x − a) + c2(x − a)2
+ … =
∞
∑
n=0
cn(x − a)n
is differentiable (and therefore continuous) on the interval (a − R, a + R) and
29

31. f′(x) = c1 + 2c2(x − a) + 3c3(x − a)2
+ … =
∞
∑
n=1
ncn(x − a)n−1
∫
f(x) dx = C + c0(x − a) + c1
(x − a)2
2
+ c2
(x − a)3
3
+ … = C +
∞
∑
n=0
cn
(x − a)n+1
n + 1
Note: The radii of convergence of these two power series is R.
Power series representation (expansion)
If f has a power series representation (expansion) at a, that is, if
f(x) =
∞
∑
n=0
cn(x − a)n
|x − a| < R
then its coefficients are given by
cn =
f (n)
(a)
n!
and the power series has the form
f(x) =
∞
∑
n=0
f (n)
(a)
n!
(x − a)n
= f(a) +
f′(a)
1!
(x − a) +
f′′(a)
2!
(x − a)2
+
f′′′(a)
3!
(x − a)3
+ . . .
which is the taylor series of the function f at a (or about a or centered at a).
30

32. Taylor series
The taylor series of a function f at a (or about a or centered at a) is
f(x) =
∞
∑
n=0
f (n)
(a)
n!
(x − a)n
= f(a) +
f′(a)
1!
(x − a) +
f′′(a)
2!
(x − a)2
+
f′′′(a)
3!
(x − a)3
+ . . .
Remainder of the taylor series
If f(x) = Tn(x) + Rn(x) where Tn is the nth-degree taylor polynomial of f at a
and
lim
n→∞
Rn(x) = 0
for |x − a| < R, then f is equal to the sum of its taylor series on the interval
|x − a| < R.
Taylor’s inequality
If
| f (n+1)
(x)| ≤ M for |x − a| ≤ d
then the remainder Rn(x) of the taylor series satisfies the inequality
|Rn(x)| ≤
M
(n + 1)!
|x − a|n+1
for |x − a| ≤ d
31

33. Maclaurin series
The maclaurin series is a specific instance of the taylor series where a = 0. In other words,
it’s just the taylor series of a function f at 0 (or about 0 or centered at 0).
f(x) =
∞
∑
n=0
f (n)
(0)
n!
(x − 0)n
=
∞
∑
n=0
f (n)
(0)
n!
xn
= f(0) +
f′(0)
1!
(x − 0) +
f′′(0)
2!
(x − 0)2
+
f′′′(0)
3!
(x − 0)3
+ . . .
= f(0) +
f′(0)
1!
x +
f′′(0)
2!
x2
+
f′′′(0)
3!
x3
+ . . .
Common maclaurin series and their radii of convergence
1
1 − x
=
∞
∑
n=0
xn
= 1 + x + x2
+ x3
+ . . . R = 1
ex
=
∞
∑
n=0
xn
n!
= 1 +
x
1!
+
x2
2!
+
x3
3!
+ … R = ∞
sin x =
∞
∑
n=0
(−1)n x2n+1
(2n + 1)!
= x −
x3
3!
+
x5
5!
−
x7
7!
+ … R = ∞
cos x =
∞
∑
n=0
(−1)n x2n
(2n)!
= 1 −
x2
2!
+
x4
4!
−
x6
6!
+ … R = ∞
32

34. tan−1
x =
∞
∑
n=0
(−1)n x2n+1
2n + 1
= x −
x3
3
+
x5
5
−
x7
7
+ R = 1
ln(1 + x) =
∞
∑
n=1
(−1)n−1 xn
n
= x −
x2
2
+
x3
3
−
x4
4
+ … R = 1
(1 + x)k
=
∞
∑
n=0
(
k
n)
xn
= 1 + kx +
k(k − 1)
2!
x2
+
k(k − 1)(k − 2)
3!
x3
+ . . . R = 1
Exponential series
ex
=
∞
∑
n=0
xn
n!
for all x
e =
∞
∑
n=0
1
n!
= 1 +
1
1!
+
1
2!
+
1
3!
+ . . .
Binomial series
If k is any real number and |x| < 1, then
(1 + x)k
=
∞
∑
n=0
(
k
n)
xn
= 1 + kx +
k(k − 1)
2!
x2
+
k(k − 1)(k − 2)
3!
x3
+ . . .
33

35. Recognizing types of series
Use this visual guide to train yourself to quickly recognize different types of series.
Sometimes the best way to recognize the type of series is to write out the first few terms
of the series and then match it to one of the types of series below.
Geometric series
2.3171717... = 2.3 +
17
103
+
17
105
+
17
107
+ . . . = 2.3 +
17
103 (
1 +
1
102
+
1
104
+ . . .
)
5 −
10
3
+
20
9
−
40
27
+ . . . = 5
[
1 −
2
3
+
(
2
3)
2
−
(
2
3)
3
+ . . .
]
4 + 3 +
9
4
+
27
16
+ . . . = 4
[
1 +
3
4
+
(
3
4)
2
+
(
3
4)
3
+ . . .
]
3 − 4 +
16
3
−
64
9
+ . . . = 3
[
1 −
4
3
+
(
4
3)
2
−
(
4
3)
3
+ . . .
]
10 − 2 + 0.4 − 0.08 + . . . = 10 [1 − 0.2 + (0.2)2
− (0.2)3
+ . . . ]
2 + 0.5 + 0.125 + 0.03125 + = 2 [1 + 0.25 + (0.25)2
+ (0.25)3
+ . . . ]
p-series
∞
∑
n=1
1
n3
=
1
13
+
1
23
+
1
33
+
1
43
+ . . .
34

36. ∞
∑
n=1
1
n
1
3
=
∞
∑
n=1
1
3
n
= 1 +
1
3
2
+
1
3
3
+
1
3
4
+ . . .
Telescoping series
∞
∑
n=1
(
1
n
−
1
n + 1)
=
(
1 −
1
2)
+
(
1
2
−
1
3)
+
(
1
3
−
1
4)
+ . . . +
(
1
n
−
1
n + 1 )
Alternating series
(−1)n
(−1)n−1
(−1)n+1
Will start with − Will start with + Will start with +
∞
∑
n=1
(−1)n
n
2n + 3
∞
∑
n=1
(−1)n−1
2n + 1
∞
∑
n=1
(−1)n+1 n2
n3 + 4
∞
∑
n=1
(−1)n n
n3 + 2
∞
∑
n=1
(−1)n−1
e
2
n
∞
∑
n=1
(−1)n+1
ne−n
∞
∑
n=1
(−1)n 3n − 1
2n + 1
∞
∑
n=1
(−1)n−1
arctan n
∞
∑
n=1
(−1)n
e−n
∞
∑
n=1
(−1)n−1
ln(n + 4)
35

37. 36