Week 11 Guide: Ratio Test, Power Series, and Taylor Series

syrupsandwich

March 28, 2025

“Math isn’t about memorizing formulas—it’s about understanding the story they tell.” — Lefty Gunplay

0.1 What’s This All About?

numbers—like 1 + 21 + 14 + · · ·. Does it stop at a number (like 2), or grow

Imagine adding an infinite list of P
forever? That’s a series, written an . The ratio test is your tool to figure this out by checking how terms
change.
Here’s the process:
an+1
- Take two terms: an (current) and an+1 (next). - Form the ratio: an (absolute value keeps it positive).
- Find the limit as n gets huge: L = limn→∞ aan+1
n
.
The rules are:
- L < 1: Converges (terms shrink, sum is finite). - L > 1 (or infinity): Diverges (terms grow or stay big). -
L = 1: Inconclusive (need another test).

0.2 Why It’s Like a Geometric Series

Think of a geometric series like 1 + 2 + 4 + 8 + · · ·. Each term doubles (ratio P n= 2), so it diverges. But
1 + 21 + 14 + · · · halves each time (ratio = 12 ), hitting 2. A geometric series ar converges if |r| < 1. The
ratio test checks if your series shrinks like this.

0.3 Let’s Try It: Step-by-Step Examples with Exponent and Factorial Properties Explained

Before we begin, recall these key properties:

• Exponents: For any real number b and integer n,

bn+1 = b · bn .

• Factorials: For any positive integer n,

(n + 1)! = (n + 1) · n!.

These properties allow us to break up and simplify expressions when using the ratio test.
P∞ 3n
Example 1: n=1 n!
Step 1: Write the nth term and the (n + 1)th term.

3n 3n+1
an = and an+1 = .
n! (n + 1)!

1
Week 11 2

Step 2: Form the ratio:

3n+1
an+1 (n+1)! 3n+1 n!
= 3n = · .
an n!
(n + 1)! 3n
Step 3: Simplify using the properties:

3n+1
• 3n : Using the exponent rule, we have

3n+1
3n+1 = 3 · 3n , so = 3.
3n

• n!
(n+1)! : Using the factorial rule,

n! 1
(n + 1)! = (n + 1) · n! thus = .
(n + 1)! n+1

Thus, the ratio becomes:

an+1 1 3
=3· = .
an n+1 n+1
Step 4: Take the limit:
3
= 0.
L = lim
n+1 n→∞

Result: Since L = 0 < 1, the series converges. Note that factorials (e.g., 5! = 120) grow much faster than
exponentials (e.g., 35 = 243) for large n.

P∞ n2
Example 2: n=1 2n
Step 1: Write the terms:
n2 (n + 1)2
an = and an+1 = .
2n 2n+1
Step 2: Form the ratio:
an+1 (n + 1)2 2n
= · 2.
an 2n+1 n
Step 3: Simplify:
2n
• Simplify the exponents: 2n+1 = 1
2 (since 2n+1 = 2 · 2n ).
• The remaining fraction becomes:
2 2
(n + 1)2
 
n+1 1
= = 1+ .
n2 n n

Thus, the ratio is:

 2
an+1 1 1
= 1+ · .
an n 2
Step 4: Take the limit:
 2
1 1 1 1
L = lim 1 + · =1· = .
n→∞ n 2 2 2
1
Result: Since 2 < 1, the series converges. Here, the exponential factor 2n grows faster than the polynomial
n2 .
P∞ 5n
Example 3: n=1 n
Step 1: Write the terms:
5n 5n+1
an = and an+1 = .
n n+1
Week 11 3

Step 2: Form the ratio:

an+1 5n+1 n
= · .
an n + 1 5n
Step 3: Simplify using exponent rules:

5n+1
• 5n = 5 (since 5n+1 = 5 · 5n ).
• The remaining factor is n
n+1 .

Thus, the ratio simplifies to:

an+1 n
=5· .
an n+1
Step 4: Take the limit:
n
L = lim 5 ·
= 5 · 1 = 5.
n+1 n→∞

Result: Since 5 > 1, the series diverges because the exponential 5n dominates, causing the terms not to
shrink fast enough.
P∞ n+1
Example 4: n=1 n
Step 1: Write the terms:
n+1 n+2
an = and an+1 = .
n n+1
Step 2: Form the ratio:
n+2
an+1 n+1 n+2 n
= n+1 = · .
an n
n+1 n+1
Step 3: Combine the fractions:
an+1 n(n + 2)
= .
an (n + 1)2
Step 4: Take the limit:
n2 + 2n
L = lim .
n→∞ n2 + 2n + 1
Since for large n the highest degree terms dominate:

n2
L ≈ lim = 1.
n→∞ n2

n+1
Result: Since L = 1, the ratio test is inconclusive. (In fact, because the terms an = n approach 1 rather
than 0, the series diverges by the Test for Divergence.)

0.4 When It’s Inconclusive

n
L = 1 means the test can’t decide. If terms don’t shrink to 0 (like n+1 → 1), it might diverge, but you need
P1
another method (e.g., compare to n ).

0.5 When to Use It

Best for series with factorials (n!) or exponentials (bn )—they simplify nicely in ratios.
Week 11 4

0.6 Practice Problems

Try these:

6n
P∞
1. n=1 n!

n4
P∞
2. n=1 3n

2n
P∞
3. n=1 n2
P∞ n
4. n=1 2n
P∞ n!
5. n=1 4n

Solutions:

6n
P∞
1. n=1 n!
n
Let’s figure out if this series converges using the ratio test. First, identify the general term: an = 6n! .
1 2
This means the first term is 61! = 6, the second is 62! = 36 2 = 18, and so on. Factorials grow fast—like
5! = 5 · 4 · 3 · 2 · 1 = 120—so let’s see if they beat 6n .
6n+1 2 3
Next, write the next term: an+1 = (n+1)! . For example, if n = 2, a2 = 62! = 18, and a3 = 63! = 216
6 =
36. n+1
6
an+1 6n+1
Now, form the ratio: an = (n+1)!
6n . To simplify, rewrite it as (n+1)! · 6n!n . The exponents: 6n+1 = 6 · 6n ,
n!
n+1
so 66n = 6. The factorials: (n + 1)! = (n + 1) · n!, so (n+1)!
n! n!
= (n+1)·n! 1
= n+1 6
. Combine: n+1 .
6 6
Take the limit: L = limn→∞ n+1 . As n gets big, the denominator grows. Try n = 10: 11 ≈ 0.545.
6 6
Try n = 100: 101 ≈ 0.059. Try n = 1000: 1001 ≈ 0.006. It’s heading to 0 because the bottom keeps
growing while 6 stays fixed. So, L = 0.
Since 0 < 1, the series converges. The factorials in the denominator make the terms shrink super fast
compared to 6n .
Answer: Converges.
P∞ n4
2. n=1 3n
n4 14
Let’s use the ratio test to see if this converges. The term is an = 3n . For n = 1, it’s 31 = 13 ; for n = 2,
24 16 n
32 = 9 ≈ 1.78. Does 3 grow fast enough to shrink this?
(n+1)4
(n+1)4 an+1 3n+1 (n+1)4 3n
Next term: an+1 = 3n+1 . So, the ratio is an = n4
= 3n+1 · n4 . Simplify the exponents:
3n
n n (n+1)4
4
4
4
3 3 1
3n+1 = 3·3n = 3 . The powers: n4 = n+1 n = 1 + n1 . So, aan+1
n
= 1 + n1 · 13 .

4
Find the limit: L = limn→∞ 1 + n1 · 13 . As n grows, n1 shrinks. For n = 10, 1 + 10 1
= 1.1,
4 1.4641 1 4 1.0406
(1.1) = 1.4641, 3 ≈ 0.488. For n = 100, 1 + 100 = 1.01, (1.01) ≈ 1.0406, 3 ≈ 0.347. It’s
approaching 14 · 13 = 1 · 31 = 13 . So, L = 13 .
Since 31 < 1, it converges. The exponential 3n outpaces n4 .
Answer: Converges.
P∞ 2n
3. n=1 n2
2n 21
Let’s use the ratio test to check if this converges. The term is an = n2 . For n = 1, it’s 12 = 2; for
2
n = 2, 222 = 44 = 1. Does n2 grow fast enough to tame 2n ?
2n+1
2n+1 n+1 2
So, the ratio is aan+1
2 2 n
Next term: an+1 = (n+1)2 . n
= (n+1)
2n = (n+1) 2 · 2n . Simplify the exponents:
n2
 2  2
2n+1 n2
2n = 2. The squares: (n+1)2 = n+1
n
. So, aan+1
n
= 2 · n
n+1 .
 2
n n
Find the limit: L = limn→∞ 2 · n+1 . As n gets big, n+1 gets close to 1. For n = 10, 10
11 ≈ 0.909,
Week 11 5

(0.909)2 ≈ 0.826, 2 · 0.826 ≈ 1.652. For n = 100, 100 2

101 ≈ 0.99, (0.99) ≈ 0.98, 2 · 0.98 ≈ 1.96. It’s heading
2
toward 2 · 1 = 2. So, L = 2.
Since 2 > 1, it diverges. The exponential 2n grows faster than n2 can handle.
Answer: Diverges.
P∞ n
4. n=1 2n
n
Let’s apply the ratio test. Term: an = 2n . Like n = 1, 21 ; n = 2, 2
4 = 0.5.
n+1
n
an+1 2n 2n
Next: an+1 = 2n+1n+1 . Ratio: an =
2n+1
n = n+1
2n+1 · 2
n . Exponents: 2n+1 = 2·2n = 1
2. Numerators:
2n
n+1 1 an+1 1

1
n = 1 + n . So, an = 1 + n · 2.
1

1 1 1.1 1
Limit: L = limn→∞ 1 + n · 2 . For n = 10, 1 + 10 = 1.1, 2 = 0.55. For n = 100, 1 + 100 = 1.01,
1.01 1 1
2 = 0.505. It goes to 2 . So, L = 2 .
1 n
Since 2 < 1, it converges. 2 grows faster than n.
Answer: Converges.
P∞ n!
5. n=1 4n
n! 1! 1 2! 2
Ratio test time. Term: an = 4n . Examples: n = 1, 41 = 4 ; n = 2, 42 = 16 = 0.125.
(n+1)!
an+1 n
Next: an+1 = (n+1)!
4n+1 . Ratio: an = 4n+1
n! = (n+1)!
4n+1 · 4n! . Factorials: (n + 1)! = (n + 1) · n!, so
4n
(n+1)! 4n 4n 1 an+1 1 n+1
n! = n + 1. Exponents: 4n+1 = 4·4n = 4 . So, an = (n + 1) · 4 = 4 .
Limit: L = limn→∞ n+1 11
4 . As n grows, this gets huge. n = 10, 4 = 2.75; n = 100, 4 =
101
25.25;
1001
n = 1000, 4 = 250.25. It’s going to infinity because n + 1 keeps increasing while 4 stays the same.
So, L = ∞.
Since ∞ > 1, it diverges. Factorials grow way faster than 4n .
Answer: Diverges.

1 Power Series: Turning Functions into Infinite Polynomials

1.1 What’s a Power Series?

A power series is an infinite sum of terms involving powers of (x − a), where a is the center:

∞
X
cn (x − a)n = c0 + c1 (x − a) + c2 (x − a)2 + c3 (x − a)3 + . . .
n=0

Each term in the series has a coefficient cn and a power (x − a)n . The coefficients cn determine how much
each term contributes to the sum.

1.2 Understanding the Coefficients cn

To identify cn in a given power series, follow these steps:

1. Compare the given series to the general form:

∞
X
cn (x − a)n
n=0

2. Identify the part that depends on x (which should be in the form of (x − a)n ) and the part that is a
coefficient (which is cn ).
P∞ n P∞
For example: - In n=0 x3n , the coefficient is cn = 31n , and the center is a = 0. - In n=0 n(x − 1)n , the
coefficient is cn = n, and the center is a = 1.
Week 11 6

1.3 Finding the Radius of Convergence

To determine for which values of x the series converges, we use the Ratio Test:

cn+1
L = lim
n→∞ cn

This tells us how the terms behave as n grows. If L exists, we find the radius of convergence R using:

1
R=
L
This means the series converges absolutely for |x − a| < R.

1.4 Step-by-Step Process for Finding R

cn+1 1
1. Identify cn . 2. Compute cn . 3. Take the limit as n → ∞. 4. Use R = L.
If L = 0, the series converges for all x (R = ∞). If L is infinite, the series only converges at x = a (R = 0).
Next, we apply this process to examples.

1.5 Examples
P∞ xn
Example 1: n=0 3n
- Here, cn = 31n , and a = 0.
- Compute the ratio:
cn+1 1/3n+1 1
= n
=
cn 1/3 3
- Limit: L = |x| · 13 , so R = 1/3
1
= 3.
P∞ n
Example 2: n=0 n(x − 1)
- Here, cn = n, and a = 1.
- Compute the ratio:
cn+1 n+1
=
cn n
n+1
- Limit: L = |x − 1| · limn→∞ n = |x − 1| · 1 = |x − 1|, so R = 1.
P∞ (x+2)n
Example 3: n=0 n!
1
- Here, cn = n! , and a = −2.
- Compute the ratio:
cn+1 1/(n + 1)! 1
= =
cn 1/n! n+1
1
- Limit: L = |x + 2| · limn→∞ n+1 = 0, so R = ∞ (converges for all x).
P∞ 2 n
Example 4: n=0 n x
2
- Here, cn = n , and a = 0.
- Compute the ratio:
cn+1 (n + 1)2
=
cn n2
(n+1)2
- Limit: L = |x| · limn→∞ n2 = |x| · 1 = |x|, so R = 1.
 Week 11 7

1.6 Interval of Convergence

To find the interval of convergence, we test the endpoints of the interval |x − a| ≤ R.

P∞ (x−2)n
Example 1: n=1 n2
Given: Center a = 2, radius of convergence R = 1.

Step 1: Start with the open interval:

The radius R = 1 means the series converges when |x − 2| < 1.
Solve: −1 < x − 2 < 1 ⇒ 1 < x < 3.
So, the open interval is (1, 3).
Step 2: Identify the endpoints:
Left endpoint: x = 2 − 1 = 1.
Right endpoint: x = 2 + 1 = 3.

Step 3: Test the left endpoint x = 1:

Substitute x = 1:
∞ ∞
X (1 − 2)n X (−1)n
2
= .
n=1
n n=1
n2

Alternating series: − 11 + 14 − 19 + · · ·.
Alternating Series Test: n12 decreases, limn→∞ 1
n2 = 0.
Converges at x = 1.
Step 4: Test the right endpoint x = 3:
Substitute x = 3:
∞ ∞
X (3 − 2)n X 1
2
= 2
.
n=1
n n=1
n
P-series with p = 2 > 1, so it converges.
Step 5: Write the interval of convergence:
Converges on (1, 3), and at x = 1 and x = 3.
Final interval: [1, 3].

Interval of convergence: [1, 3].

P∞ (x+3)n
Example 2: n=1 n·5n
Given: Center a = −3, radius of convergence R = 5.

Step 1: Start with the open interval:

The radius R = 5 means convergence when |x + 3| < 5.
Solve: −5 < x + 3 < 5 ⇒ −8 < x < 2.
Open interval: (−8, 2).
Step 2: Identify the endpoints:
Left endpoint: x = −3 − 5 = −8.
Right endpoint: x = −3 + 5 = 2.
 Week 11 8

Step 3: Test the left endpoint x = −8:

Substitute x = −8:
∞ ∞ ∞
X (−8 + 3)n X (−5)n X (−1)n
n
= n
= .
n=1
n·5 n=1
n·5 n=1
n

Alternating harmonic series: −1 + 21 − 13 + · · ·.

Alternating Series Test: n1 decreases, limn→∞ n1 = 0.
Converges at x = −8.
Step 4: Test the right endpoint x = 2:
Substitute x = 2:
∞ ∞ ∞
X (2 + 3)n X 5n X 1
n
= n
= .
n=1
n · 5 n=1
n · 5 n=1
n

Harmonic series, diverges (p-series with p = 1 ≤ 1).

Step 5: Write the interval of convergence:

Converges on (−8, 2), converges at x = −8, diverges at x = 2.
Final interval: [−8, 2).

Interval of convergence: [−8, 2).

1.7 Practice Problems

xn
P∞
1. n=0 4n
P∞
2. n=0 n!(x − 3)n
P∞ (x+1)n
3. n=0 n2n

xn
P∞
4. n=1 n5n

Solutions:

xn
P∞
1. n=0 4n P∞ xn
We need to find where this power series converges—its radius and interval. Rewrite it: n=0 4n =
x n
P∞
P n
n=0 4 . The general form is c n (x − a) , so here a = 0 (no shift), and c n = 1 (no extra factors).
cn+1 1 cn+1
Use the ratio test for the coefficients: cn = 1 = 1. The limit is limn→∞ cn = 1. The radius is
1 1 1

n
cn x4 , cn
= 1, ratio still 1, but L = x4 < 1,
P
R= = = 1. But adjust for the
lim 1 4n : rewrite as
so |x| < 4, R = 4.
P 4n P (−4)n
= (−1)n ,
P P
Interval: −4 < x < 4. Endpoints: x = 4, 4n = 1, diverges (all 1s). x = −4, 4n
alternates 1, -1, diverges. So, (−4, 4).
Answer: R = 4, (−4, 4).
P∞ n
2. n=0 n!(x − 3)
cn (x − a)n , a = 3, cn = n! (e.g., 0! = 1, 1! = 1, 2! = 2).
P
Let’s find the radius and interval. Form:
cn+1 (n+1)! 1
Ratio: cn = n! = n + 1. Limit: limn→∞ |n + 1| = ∞. Radius: R = ∞ = 0. This means it only
converges at x =P3.
x = 3, P n!(3 − 3)n = n! · 0n = 1 + 0 + 0 + · · · = 1, converges. Any x ̸= 3, like x = 4,
P
Check:
n
P
n!(4 − 3) = n!, diverges (factorials explode). So, interval is just {3}.
Answer: R = 0, {3}.
Week 11 9

P∞ (x+1)n
3. n=0 n2n
P (x+1)n 1
cn (x − a)n . Notice x + 1 = x − (−1), so a = −1, cn =
P
Rewrite: n·2n = n2n .
1
cn+1 n n
(n+1)2n+1 n2 n 2 n
Ratio: cn = 1 = (n+1)2n+1 = n+1 · 2·2n = n+1 · 12 . Limit: limn→∞ n
n+1 · 1
2 = 1· 1
2 = 1
2.
n2n
1
Radius: R = 1 = 2.
2
P (−3+1)n P (−2)n
Interval: |x + 1| < 2, so −2 < x + 1 < 2, −3 < x < 1. Endpoints: x = −3, n2n = n2n =
P (−1)n P (1+1)n P1
n , converges (alternating harmonic). x = 1, n2n = n , diverges. So, [−3, 1).
Answer: R = 2, [−3, 1).
P∞ xn
4. n=1 n5nP
xn
P 1 x
n
Rewrite: n5n = n 5 , so a = 0, cn = n1 .
1
cn+1 n n
Ratio: cn = n+1
1 = = 1. L = |x| · 51 , R = 5.
n+1 , limn→∞ n+1
n
P 5n P1 P (−5)n P (−1)n
Interval: −5 < x < 5. Endpoints: x = 5, n5n = n , diverges. x = −5, n5n = n ,
converges. So, [−5, 5).
Answer: R = 5, [−5, 5).

2 Which Test to Pick?

2.1 Your Toolkit

1
- n! or bn : Ratio test. - Power series: Ratio test for R. - np : p-series.

2.2 Examples

n!
P
Example 1: 5n Ratio, L = ∞, diverges.
1
P
Example 2: n3 p-series, p = 3, converges.
P 2n
Example 3: n! Ratio, L = 0, converges.
n
Ratio, L = 31 , converges.
P
Example 4: 3n

2.3 Practice Problems

P 4n
1. n3
1
P
2. n4
P n2
3. 2n

n!
P
4. nn

Solutions:
P 4n
1. n3
Let’s pick a test. This has an exponential 4n and a polynomial n3 , so the ratio test is perfect. Term:
n
an = 4n3 . For n = 1, 41 = 4; n = 2, 16
8 = 2.
4n+1
4n+1 an+1 (n+1)3 4n+1 n3 4n+1 n3
Next: an+1 = (n+1)3 . Ratio: an = 4n = (n+1)3 · 4n . Exponents: 4n = 4. Cubes: (n+1)3 . So,
n3
4n3
(n+1)3 .
4n3
Limit: L = limn→∞ (n+1)3 . Expand (n + 1)3 = n3 + 3n2 + 3n + 1. For large n, n3 dominates, so
Week 11 10

4n3 4n3 4 4
n3 +3n2 +3n+1 ≈ n3 = 4. Divide by n3 : 3
1+ n + n32 + n13
→ 1 = 4. L = 4.
n 3
Since 4 > 1, it diverges. 4 grows faster than n .
Answer: Diverges.
P 1
2. n4 P 1
Which test? This is np with p = 4. For p-series, if p > 1, it converges; if p ≤ 1, it diverges. Here,
p = 4, and 4 > 1. Terms like 114 = 1, 214 = 16
1
, shrink fast. So, it converges.
Answer: Converges.
P n2
3. 2n
n2 (n+1)2
Exponential 2n and polynomial n2 suggest ratio test. Term: an = 2n . Next: an+1 = 2n+1 . Ratio:
(n+1)2
(n+1)2 2n (n+1)2
2n+1
n2
= 2n+1 · n2 = n2 · 12 .
2n
n2 +2n+1 2 1 an+1 2 1
· 12 . Limit:


Expand: (n + 1)2 = n2 + 2n + 1, so n2 = 1+ n + n2 . Then, an = 1+ n + n2
limn→∞ 1 + n2 + n12 · 12 = 1 · 12 = 12 .

Since 21 < 1, it converges.

Answer: Converges.
P n!
4. nn
n! (n+1)!
Factorial and exponential-like nn , so ratio test. Term: an = nn . Next: an+1 = (n+1)n+1 . Ratio:
(n+1)!
(n+1)! n
(n+1)n+1 n
n! = (n+1)n+1 · n! .
nn  n
nn nn
Factorials: (n+1)! = n + 1. Powers: n+1 = n = 1
· n
. So, aan+1 1
= (n + 1) · n+1 ·
 n  n! n (n+1) (n+1)·(n+1) n+1 n+1 n

n n
n+1 = n+1 .
 n  n
n
n
Limit: L = limn→∞ n+1 = limn→∞ 1+1 1 = 1e ≈ 0.368 (since 1 + n1 → e).
n
Since 1e < 1, it converges.
Answer: Converges.

3 Playing with Power Series

3.1 What Can You Do?

Add, subtract, differentiate, or multiply power series within the smallest radius R.
- Add: Combine like terms.
- Differentiate: Adjust powers/coefficients.
- Multiply: Distribute terms.

3.2 Examples

Example 1: ex + e−x

P∞ xn
P∞ (−1)n xn
1. ex = n=0 n! , e−x = n=0 n! .
P∞ xn (1+(−1)n )
2. n=0 n! .
2x2m
P∞
3. Even n = 2m: 2, odd: 0. So, m=0 (2m)! .

4. R = ∞.
Week 11 11

∞
X 2x2m
.
m=0
(2m)!

xn
P
Example 2: Differentiate

P∞ 1
1. n=0 xn = 1−x , |x| < 1.
d n
P∞
2. dx (x ) = nxn−1 , so n=1 nxn−1 .
1
3. Matches (1−x)2 .

4. R = 1.
∞
X
nxn−1 .
n=1

Example 3: sin x + cos x

P∞ (−1)n x2n+1 P∞ (−1)n x2n

1. sin x = n=0 (2n+1)! , cos x = n=0 (2n)! .

P∞ (−1)n x2n+1 P∞ (−1)n x2n

2. Combine: n=0 (2n+1)! + n=0 (2n)! .

3. No overlap, keep as is.

4. R = ∞.
∞ ∞
X (−1)n x2n+1 X (−1)n x2n
+ .
n=0
(2n + 1)! n=0
(2n)!

xn by x
P
Example 4: Multiply

P∞
1. n=0 xn , |x| < 1.
P∞
2. x · (1 + x + x2 + · · · ) = n=0 xn+1 .

3. Starts at x, correct pattern.

4. R = 1.
∞
X
xn+1 .
n=0

3.3 Practice Problems

P xn
1. Differentiate n! .
P n
x and (−x)n .
P
2. Add

3. cos x − 1.
P xn
4. x · n .

Solutions:
Week 11 12

P xn
1. Differentiate
P∞n! xn
We’re given n=0 n! , which is the series for ex . Let’s differentiate it term by term. Write it out:
2 3
1 + x + x2! + x3! + · · ·.  2  3  4
d d d x 2x x d x 3x2 x2 d x
Differentiate each term: dx (1) = 0, dx (x) = 1, dx 2! = 2! = ,
1! dx 3! = 3! = ,
2! dx 4! =
4x3 x3 d xn nxn−1 xn−1


4! = 3! . Notice the pattern: dx n! = n! = (n−1)! (for n ≥ 1), and the n = 0 term becomes 0.
0 1 2 P∞ xn−1
Adjust indices: for n = 1, x0! = 1; n = 2, x1! = x; n = 3, x2! . So, the series is n=1 (n−1)! . Let
P∞ xk x
k = n − 1, when n = 1, k = 0, when n = 2, k = 1, so k=0 k! = e .
P∞ xn
Answer: n=0 n! .
P n P n
2. Add x and
P∞ (−x) P∞
First series: n=0 x = 1+x+x2 +x3 +· · ·, R = 1. Second: n=0 (−x)n = 1+(−x)+(−x)2 +(−x)3 =
n

1 − x + x2 − x3 + · · ·, R = 1.
Add term by term: n = 0, 1 + 1 = 2; n = 1, x + (−x) = 0; n = 2, x2 + x2 = 2x2P ; n = 3, x3 + (−x3 ) = 0.
∞
Pattern: even
P∞n terms double, odd cancel. So, 2 + 0 + 2x + 0 + 2x + · · · = n=0 2x2n .
2 4
2n
Answer: n=0 2x .

3. cos x − 1P n 2n
∞ 2 4 2 4 2 4
cos x = n=0 (−1) x
(2n)! = 1 − x2! + x4! − · · ·. Subtract 1: 1 − x2! + x4! − · · · − 1 = − x2! + x4! − · · ·. Adjust:
P∞ (−1)n x2n
n=1 (2n)! .
P∞ (−1)n x2n
Answer: n=1 (2n)! .
P xn
4. x · n
P∞ xn x2 x3 2 x2 x3 x3 x4
Series: n=1 n = x + 2 + 3 + · · ·. Multiply by x: x · x = x , x · 2 = 2 , x· 3 = 3 . Pattern:
xn xn+1 ∞ xn+1
P
x · n = n . So, n=1 n .
P∞ xn+1
Answer: n=1 n .

4 Taylor Series: Function Magic

4.1 What’s a Taylor Series?

A Taylor series writes a function as an infinite sum using its derivatives at a point a:
∞
X f (n) (a)
f (x) = (x − a)n .
n=0
n!

Steps: Find derivatives at a, plug into formula, simplify.

4.2 Key Series

xn
P∞
- ex = n=0 n! (all x).
P∞ n 2n+1
- sin x = n=0 (−1) x
(2n+1)! (all x).
P∞ (−1)n x2n
- cos x = n=0 (2n)! (all x).
1
P∞ n
- 1−x = n=0 x (|x| < 1).

4.3 Building from Scratch

Example 1: ex at a = 0
Week 11 13

1. All derivatives ex , at 0: f (n) (0) = 1.

P∞ 1 n
2. n=0 n! x .

∞
X xn
.
n=0
n!

Example 2: sin x at a = 0

1. Derivatives cycle: 0, 1, 0, -1 at x = 0.
P∞ (−1)n x2n+1
2. Odd terms, alternating: n=0 (2n+1)! .

∞
X (−1)n x2n+1
.
n=0
(2n + 1)!

4.4 Using Known Series

Example 3: cos(x2 )

P∞ (−1)n x2n
1. cos x = n=0 (2n)! .

P∞ (−1)n (x2 )2n P∞ (−1)n x4n

2. Replace x with x2 : n=0 (2n)! = n=0 (2n)! .

∞
X (−1)n x4n
.
n=0
(2n)!

1
Example 4: 1−3x

1
P∞
1. 1−x = n=0 xn , |x| < 1.
P∞ P∞
2. Replace x with 3x: n=0 (3x)
n
= n=0 3n xn , |x| < 13 .
∞
X
3n xn .
n=0

Now its test time

Week 11 14

Test 1: Ratio Test, Power Series, and Taylor Series

Questions

1. (Ratio Test) Determine whether the series

∞
X 4n
n=1
n!

converges or diverges.
2. (Power Series) Find the radius and interval of convergence for the power series
∞
X (x − 2)n
.
n=0
3n

3. (Taylor Series) Find the Taylor series expansion for ln(1 + x) about a = 0 and state
its interval of convergence.
4. (Ratio Test) Determine the convergence or divergence of the series
∞
X n
.
n=1
3n

5. (Function Identification) For the series

∞
X x2n
(−1)n ,
n=0
(2n)!

find its radius of convergence and identify the function it represents.

Solutions

1. Solution: For the series

4n
an = ,
n!
we apply the ratio test:
an+1 4n+1 n! 4
= · n = .
an (n + 1)! 4 n+1
Taking the limit as n → ∞:
4
L = lim = 0.
n→∞ n + 1

Since L < 1, the series converges by the ratio test.

2. Solution: The power series
∞
X (x − 2)n
n=0
3n
Week 11 15

can be rewritten as ∞  n
X x−2
.
n=0
3
x−2
This is a geometric series with ratio r = 3
. The series converges when

x−2
<1 =⇒ |x − 2| < 3.
3
Thus, the radius of convergence is R = 3 and the interval of convergence is
(2 − 3, 2 + 3) = (−1, 5).
Endpoint Analysis:
• For x = −1: r = −1−2
(−1)n (divergent).
P
3
= −1 leading to the series
• For x = 5: r = 5−2
P n
3
= 1 yielding 1 (divergent).
Therefore, the interval is (−1, 5) with endpoints excluded.
3. Solution: The Taylor series for ln(1 + x) about a = 0 is a known expansion:
∞
X xn
ln(1 + x) = (−1)n+1 , for |x| < 1.
n=1
n

Hence, the series converges for |x| < 1.

4. Solution: Consider n
an = .
3n
Applying the ratio test:
an+1 (n + 1)/3n+1 n+1
= n
= .
an n/3 3n
Taking the limit as n → ∞:
n+1 1
L = lim = .
n→∞ 3n 3
Since L < 1, the series converges.
5. Solution: The series ∞
X x2n
(−1)n
n=0
(2n)!
is recognized as the Maclaurin series for cos x. Moreover, applying the ratio test (or
using standard results) shows that the radius of convergence is infinite. Thus, the series
converges for all x and represents the function cos x.
Week 11 16

Test 2: Convergence and Series Manipulation

Questions

1. (Ratio Test) Determine whether the series

∞
X 2n
n=1
n2

converges or diverges.
2. (Power Series) Find the radius of convergence for the power series
∞
X
n(x − 1)n .
n=0

3. (Taylor Series) Derive the Taylor series expansion for e−x about a = 0 and state its
interval of convergence.
4. (Power Series) Determine the interval of convergence for the series
∞
X (x + 3)n
.
n=1
n · 5n

5. (Taylor Series) Find the Taylor series expansion for cos x at a = 0 and state its radius
of convergence.

Solutions

1. Solution: Let
2n
an = .
n2
Then, the ratio test yields:
an+1 2n+1 n2 n2
= · = 2 · .
an (n + 1)2 2n (n + 1)2
Taking the limit as n → ∞:
n2
L = lim 2 · = 2 · 1 = 2.
n→∞ (n + 1)2
Since L > 1, the series diverges by the ratio test.
2. Solution: For the power series
∞
X
n(x − 1)n ,
n=0
Week 11 17

the general term is an = n(x − 1)n . Applying the ratio test:

an+1 (n + 1)(x − 1)n+1 n+1
= n
= |x − 1|.
an n(x − 1) n
n+1
As n → ∞, n
→ 1; hence, the series converges when
|x − 1| < 1.
Thus, the radius of convergence is R = 1 and the interval of convergence is (0, 2)
(endpoint testing is required to determine convergence at x = 0 and x = 2).
3. Solution: The Taylor series for ex about 0 is
∞
x
X xn
e = .
n=0
n!

Replacing x with −x, we obtain:

∞ ∞
X (−x)n X (−1)n xn
e−x = = .
n=0
n! n=0
n!

This series converges for all x (its radius of convergence is infinite).

4. Solution: Rewrite the series as
∞  n
X 1 x+3
.
n=1
n 5

Convergence requires:
x+3
<1 =⇒ |x + 3| < 5.
5
Thus, the radius of convergence is R = 5 and the interval (before testing endpoints) is

(−8, 2).
(Endpoint analysis should be performed to check for conditional convergence, but typ-
ically the endpoints do not converge absolutely.)
5. Solution: The Taylor series expansion for cos x at a = 0 is given by
∞
X (−1)n x2n
cos x = .
n=0
(2n)!

This series is known to converge for all x; hence, its radius of convergence is infinite.