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1. Convergence of series; the nth-term test

Warm-up

Three quick questions from earlier lessons.

Discover

$\displaystyle\sum_{n=1}^{\infty} \dfrac{n}{n + 1} = \tfrac{1}{2} + \tfrac{2}{3} + \tfrac{3}{4} + \dotsb$ . The terms get close to 1. Can the sum be a finite number?

No
Yes, since each term is less than 1

Key idea

A series converges when its partial sums settle.

\displaystyle\sum a_n

converges to

S

if the partial sums

S_N = a_1 + \dotsb + a_N

tend to

S

. nth-term test: if

a_n \not\to 0

, the series diverges. If

a_n \to 0

, the test says nothing (∑ 1/n diverges).

Worked example

$\displaystyle\sum_{n=1}^{\infty} \left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right)$ , a telescoping sum.

S_N = 1 - \dfrac{1}{N + 1}

Terms cancel in pairs.

S_N \to 1

N → ∞.

The series converges to 1

So:

Your turn, guided

Your turn: $\displaystyle\sum \dfrac{3n}{n + 2}$ . The limit of the terms is…

3

, so the series diverges

3n/(n + 2) → 3 ≠ 0: diverges.

Practice ★

Does $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 6n}{n + 3}$ converge?
1. No: the terms tend to 6 ≠ 0, so it diverges (nth-term test)
2. Yes, because each term is less than 6
3. The nth-term test cannot decide
Does $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 5n}{n + 8}$ converge?
1. No: the terms tend to 5 ≠ 0, so it diverges (nth-term test)
2. Yes, because each term is less than 5
3. The nth-term test cannot decide

Find $\displaystyle\sum_{n=1}^{\infty} \left(\dfrac{ 9 }{n} - \dfrac{ 9 }{n + 1}\right)$ .
Find $\displaystyle\sum_{n=1}^{\infty} \left(\dfrac{ 3 }{n} - \dfrac{ 3 }{n + 1}\right)$ .

Watch out!

Practice ★★

$\displaystyle\sum_{n=1}^{\infty} \dfrac{ 4 }{n}$ : its terms tend to 0. What does that tell you?
1. Nothing by itself: this series in fact diverges (harmonic series)
2. The series converges
3. The sum is 0

Exit ticket

Three questions to check you have got it.

Does $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 6n}{n + 3}$ converge?
1. No: the terms tend to 6 ≠ 0, so it diverges (nth-term test)
2. Yes, because each term is less than 6
3. The nth-term test cannot decide
Does $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 5n}{n + 8}$ converge?
1. No: the terms tend to 5 ≠ 0, so it diverges (nth-term test)
2. Yes, because each term is less than 5
3. The nth-term test cannot decide
Does $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 5n}{n + 2}$ converge?
1. No: the terms tend to 5 ≠ 0, so it diverges (nth-term test)
2. Yes, because each term is less than 5
3. The nth-term test cannot decide

Summary card

Convergence

Converges: the partial sums have a limit.
$a_n \not\to 0$ : diverges. $a_n \to 0$ : test another way.

Watch out: the harmonic series diverges.

2. The integral test and p-series

Warm-up

Three quick questions from earlier lessons.

Discover

Draw rectangles of width 1 and heights $\tfrac{1}{n^2}$ from $n = 2$ on. They fit under $y = \tfrac{1}{x^2}$ from 1 to ∞, whose area is 1. So is $\displaystyle\sum_{n=2}^{\infty} \tfrac{1}{n^2}$ finite?

Yes, less than 1
No, there are infinitely many terms

Key idea

A series and an integral share their fate.

Integral test: if

f

is positive, continuous and decreasing with

a_n = f(n)

, then

\sum a_n

and

\displaystyle\int_1^{\infty} f(x)\,dx

both converge or both diverge. p-series:

\displaystyle\sum \dfrac{1}{n^p}

converges exactly when

p > 1

Worked example

Does $\displaystyle\sum \dfrac{1}{n^{1.1}}$ converge? And $\displaystyle\sum \dfrac{1}{n^{0.9}}$ ?

\displaystyle\sum \dfrac{1}{n^{1.1}}

converges

p = 1.1 > 1.

\displaystyle\sum \dfrac{1}{n^{0.9}}

diverges

p = 0.9 ≤ 1.

Your turn, guided

Your turn: the p of $\displaystyle\sum \dfrac{1}{n\sqrt{n}}$ .

p = \tfrac{3}{2}

: converges

n · n^{1/2} = n^{3/2}.

Practice ★

Which series converge? Tick all that do.
1. $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 1 }{n^2}$
2. $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 1 }{n^{3/2} }$
3. $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 1 }{\sqrt{n} }$
4. $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 1 }{n}$
Which series converge? Tick all that do.
1. $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 4 }{n^2}$
2. $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 4 }{n^{3/2} }$
3. $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 4 }{\sqrt{n} }$
4. $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 4 }{n}$

Watch out!

Practice ★★

Use the integral test on $\displaystyle\sum_{n=1}^{\infty} 7n e^{-n^2}$ .
1. It converges: the integral of 7xe^(−x²) from 1 to ∞ is finite
2. It diverges, because n grows
3. The integral test does not apply

The integral test bounds $\displaystyle\sum_{k=2}^{\infty} \dfrac{1}{k^{ 6 } }$ by $\displaystyle\int_1^{\infty} \dfrac{dx}{x^{ 6 } }$ . Find that integral.
The integral test bounds $\displaystyle\sum_{k=2}^{\infty} \dfrac{1}{k^{ 2 } }$ by $\displaystyle\int_1^{\infty} \dfrac{dx}{x^{ 2 } }$ . Find that integral.

Exit ticket

Three questions to check you have got it.

Which series converge? Tick all that do.
1. $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 1 }{n^2}$
2. $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 1 }{n^{3/2} }$
3. $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 1 }{\sqrt{n} }$
4. $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 1 }{n}$
Which series converge? Tick all that do.
1. $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 4 }{n^2}$
2. $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 4 }{n^{3/2} }$
3. $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 4 }{\sqrt{n} }$
4. $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 4 }{n}$
Which series converge? Tick all that do.
1. $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 8 }{n^2}$
2. $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 8 }{n^{3/2} }$
3. $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 8 }{\sqrt{n} }$
4. $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 8 }{n}$

Summary card

Integral test, p-series

Positive decreasing terms: compare with $\int_1^{\infty} f$ .
$\sum 1/n^p$ : converges iff $p > 1$ .

Watch out: $p = 1$ diverges.

3. Comparison tests

Warm-up

Three quick questions from earlier lessons.

Discover

$0 < \dfrac{1}{n^2 + 5} < \dfrac{1}{n^2}$ and $\sum \dfrac{1}{n^2}$ converges. What can you say about $\sum \dfrac{1}{n^2 + 5}$ ?

It converges
Nothing

Key idea

Compare with a series you know.

Direct comparison (

0 \le a_n \le b_n

): if

\sum b_n

converges, so does

\sum a_n

; if

\sum a_n

diverges, so does

\sum b_n

. Limit comparison: if

a_n, b_n > 0

and

a_n/b_n \to c

with

0 < c < \infty

, both series converge or both diverge.

Worked example

$\displaystyle\sum \dfrac{3n + 2}{n^3 - n + 4}$ .

It behaves like

\dfrac{3}{n^2}

Leading terms: 3n/n³.

Limit comparison with

\sum \dfrac{1}{n^2}

Ratio with 1/n² → 3.

Converges

p = 2 > 1.

Your turn, guided

Your turn: $\displaystyle\sum \dfrac{n}{n^2 + 1}$ behaves like $\sum \dfrac{1}{n^{p}}$ with…

p = 1

: diverges

n/n² = 1/n: diverges.

Practice ★

Does $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^2 + 7}$ converge?
1. Yes: it is smaller than the convergent ∑ 1/n²
2. No: it is smaller than the divergent ∑ 1/n
3. No test applies

Does $\displaystyle\sum_{n=1}^{\infty} \dfrac{2n + 1}{n^3 + 8}$ converge?
1. Yes: it behaves like ∑ 2/n², a convergent p-series
2. No: the top grows
3. No: it behaves like ∑ 1/n

Watch out!

Practice ★★

Does $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n} + 8}$ converge?
1. No: it behaves like ∑ 1/√n, which diverges (limit comparison)
2. Yes: it is smaller than ∑ 1/√n
3. Yes: its terms tend to 0
Does $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n} + 6}$ converge?
1. No: it behaves like ∑ 1/√n, which diverges (limit comparison)
2. Yes: it is smaller than ∑ 1/√n
3. Yes: its terms tend to 0

Exit ticket

Three questions to check you have got it.

Does $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^2 + 7}$ converge?
1. Yes: it is smaller than the convergent ∑ 1/n²
2. No: it is smaller than the divergent ∑ 1/n
3. No test applies
Does $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^2 + 8}$ converge?
1. Yes: it is smaller than the convergent ∑ 1/n²
2. No: it is smaller than the divergent ∑ 1/n
3. No test applies
Does $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^2 + 9}$ converge?
1. Yes: it is smaller than the convergent ∑ 1/n²
2. No: it is smaller than the divergent ∑ 1/n
3. No test applies

Summary card

Comparison tests

Below convergent: converges. Above divergent: diverges.
Same leading behavior: limit comparison.

Watch out: below a divergent series tells you nothing.

4. Alternating series and the error bound

Warm-up

Three quick questions from earlier lessons.

Discover

$1 - \tfrac{1}{2} + \tfrac{1}{3} - \tfrac{1}{4} + \dotsb$ : the partial sums go $1, 0.5, 0.83, 0.58, 0.78, \dots$ . Do they settle?

Yes
No, because $\sum \tfrac{1}{n}$ diverges

Key idea

Alternating signs help convergence, and give an error bound.

Alternating series test: if

b_n > 0

decreases to 0, then

\sum (-1)^{n+1} b_n

converges, and

|S - S_N| \le b_{N+1}

(the first term left out). If

\sum |a_n|

converges, the series converges absolutely; if only

\sum a_n

does, conditionally.

Worked example

$\displaystyle\sum \dfrac{(-1)^{n+1}}{n^3}$ with 3 terms.

S_3 = \tfrac{197}{216} \approx 0.912

1 − 1/8 + 1/27.

|S - S_3| \le \tfrac{1}{64}

First omitted term.

Absolutely convergent

∑1/n³ converges.

Your turn, guided

Your turn: the error bound for $\displaystyle\sum \dfrac{(-1)^{n+1}}{n}$ after 4 terms.

\tfrac{1}{5}

b₅.

Practice ★

Does $\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1} 7 }{n}$ converge?
1. Yes, conditionally: the terms alternate, shrink and tend to 0
2. No: ∑ 7/n diverges
3. Yes, absolutely

Find the partial sum $S_3$ of $\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1} 1 }{n}$ .
Find the partial sum $S_3$ of $\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1} 7 }{n}$ .

Watch out!

Practice ★★

$S = \displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1} }{n^2}$ is approximated by its first 8 terms. Give the alternating-series bound on the error.
$S = \displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1} }{n^2}$ is approximated by its first 2 terms. Give the alternating-series bound on the error.

Boss problem. $S = \displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1} 1 }{n^2}$ .

Exit ticket

Three questions to check you have got it.

$S = \displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1} }{n^2}$ is approximated by its first 8 terms. Give the alternating-series bound on the error.
$S = \displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1} }{n^2}$ is approximated by its first 2 terms. Give the alternating-series bound on the error.
$S = \displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1} }{n^2}$ is approximated by its first 6 terms. Give the alternating-series bound on the error.

Summary card

Alternating series

$b_n$ decreasing to 0: converges.
Error after $N$ terms $\le b_{N+1}$ .
Absolute vs conditional: test $\sum |a_n|$ .

Watch out: the bound uses the first omitted term.

5. The ratio test; absolute and conditional convergence

Warm-up

Three quick questions from earlier lessons.

Discover

For $\displaystyle\sum \dfrac{2^n}{n!}$ , each term is the previous one times $\dfrac{2}{n + 1}$ . For large $n$ that factor is tiny. Does the series converge?

Yes
No, $2^n$ grows

Key idea

Compare with a geometric series through the ratio of terms.

Ratio test:

L = \displaystyle\lim_{n \to \infty} \left|\dfrac{a_{n+1}}{a_n}\right|

. If

L < 1

\sum a_n

converges absolutely; if

L > 1

, it diverges; if

L = 1

, the test is inconclusive.

Worked example

$\displaystyle\sum \dfrac{n^2}{3^n}$ .

\dfrac{a_{n+1}}{a_n} = \dfrac{(n + 1)^2}{n^2} \cdot \dfrac{1}{3}

Divide consecutive terms.

L = \tfrac{1}{3}

(n + 1)²/n² → 1.

Converges absolutely

L < 1.

Your turn, guided

Your turn: $L$ for $\displaystyle\sum \dfrac{5^n}{n}$ .

L = 5

5 · n/(n + 1) → 5 > 1: diverges.

Practice ★

For $a_n = \dfrac{n}{ 8^n}$ , find $L = \displaystyle\lim_{n \to \infty} \left|\dfrac{a_{n+1} }{a_n}\right|$ .
For $a_n = \dfrac{n}{ 4^n}$ , find $L = \displaystyle\lim_{n \to \infty} \left|\dfrac{a_{n+1} }{a_n}\right|$ .
For $a_n = \dfrac{n}{ 5^n}$ , find $L = \displaystyle\lim_{n \to \infty} \left|\dfrac{a_{n+1} }{a_n}\right|$ .

Watch out!

Practice ★★

For $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^{ 2 } }$ , the ratio test gives $L = 1$ . What does it tell you?
1. Nothing: the ratio test is inconclusive when L = 1
2. The series converges
3. The series diverges

How does $\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^n 1 }{n^2}$ converge?
1. Absolutely: ∑ 1/n² converges
2. Only conditionally
3. It diverges

Exit ticket

Three questions to check you have got it.

For $a_n = \dfrac{n}{ 8^n}$ , find $L = \displaystyle\lim_{n \to \infty} \left|\dfrac{a_{n+1} }{a_n}\right|$ .
For $a_n = \dfrac{n}{ 4^n}$ , find $L = \displaystyle\lim_{n \to \infty} \left|\dfrac{a_{n+1} }{a_n}\right|$ .
For $a_n = \dfrac{n}{ 5^n}$ , find $L = \displaystyle\lim_{n \to \infty} \left|\dfrac{a_{n+1} }{a_n}\right|$ .

Summary card

Ratio test

$L = \lim |a_{n+1}/a_n|$ : $L < 1$ converges, $L > 1$ diverges, $L = 1$ unknown.

Watch out: use it for factorials and powers; not for p-series.

6. Choosing a convergence test

Warm-up

Three quick questions from earlier lessons.

Discover

$\displaystyle\sum \dfrac{n}{2n + 1}$ : before any clever test, what should you check first?

Whether the terms tend to 0
The ratio test

Key idea

A short checklist.

a_n \not\to 0

: diverges. 2. Geometric or p-series: read off. 3. Alternating: alternating series test. 4. Factorials or

c^n

: ratio test. 5. Rational or root expressions: compare with a p-series. 6. Easy integral: integral test.

Worked example

Choose tests for three series.

\displaystyle\sum \dfrac{n!}{10^n}

: ratio test,

L = \infty

, diverges

Factorial.

\displaystyle\sum \dfrac{(-1)^n}{\ln(n + 1)}

: alternating series test, converges

Signs alternate, terms decrease to 0.

\displaystyle\sum \dfrac{1}{n^2 + n}

: comparison, converges

Leading terms 1/n².

Your turn, guided

Your turn: $\displaystyle\sum \left(\tfrac{2}{3}\right)^n$ is geometric with ratio…

r = \tfrac{2}{3}

|r| < 1: converges.

Practice ★

Which test suits $\displaystyle\sum_{n=1}^{\infty} \dfrac{n!}{ 9^n}$ best?
1. The ratio test
2. Comparison with a p-series
3. The alternating series test

Which test settles $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^{ 6 } }$ fastest?
1. The p-series test: p = 6 > 1, so it converges
2. The nth-term test
3. The ratio test

Watch out!

Practice ★★

Which test shows that $\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^n 5 }{\sqrt{n} }$ converges?
1. The alternating series test
2. The p-test with p = ½
3. The ratio test
Which test shows that $\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^n 6 }{\sqrt{n} }$ converges?
1. The alternating series test
2. The p-test with p = ½
3. The ratio test

Exit ticket

Three questions to check you have got it.

Which test suits $\displaystyle\sum_{n=1}^{\infty} \dfrac{n!}{ 9^n}$ best?
1. The ratio test
2. Comparison with a p-series
3. The alternating series test
Which test suits $\displaystyle\sum_{n=1}^{\infty} \dfrac{n!}{ 8^n}$ best?
1. The ratio test
2. Comparison with a p-series
3. The alternating series test
Which test suits $\displaystyle\sum_{n=1}^{\infty} \dfrac{n!}{ 4^n}$ best?
1. The ratio test
2. Comparison with a p-series
3. The alternating series test

Summary card

Choosing a test

nth-term → known series → alternating → ratio → comparison → integral.

Watch out: a divergent absolute series does not make an alternating series diverge.

Answers: Infinite series and convergence tests

1. Convergence of series; the nth-term test

No: the terms tend to 6 ≠ 0, so it diverges (nth-term test)
No: the terms tend to 5 ≠ 0, so it diverges (nth-term test)

Nothing by itself: this series in fact diverges (harmonic series)

No: the terms tend to 6 ≠ 0, so it diverges (nth-term test)
No: the terms tend to 5 ≠ 0, so it diverges (nth-term test)
No: the terms tend to 5 ≠ 0, so it diverges (nth-term test)

2. The integral test and p-series

$\displaystyle\sum_{n=1}^{\infty} \dfrac{ 1 }{n^2}$ · $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 1 }{n^{3/2} }$
$\displaystyle\sum_{n=1}^{\infty} \dfrac{ 4 }{n^2}$ · $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 4 }{n^{3/2} }$

It converges: the integral of 7xe^(−x²) from 1 to ∞ is finite

$\frac{1}{5}$
$1$

$\displaystyle\sum_{n=1}^{\infty} \dfrac{ 1 }{n^2}$ · $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 1 }{n^{3/2} }$
$\displaystyle\sum_{n=1}^{\infty} \dfrac{ 4 }{n^2}$ · $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 4 }{n^{3/2} }$
$\displaystyle\sum_{n=1}^{\infty} \dfrac{ 8 }{n^2}$ · $\displaystyle\sum_{n=1}^{\infty} \dfrac{ 8 }{n^{3/2} }$

3. Comparison tests

Yes: it is smaller than the convergent ∑ 1/n²

Yes: it behaves like ∑ 2/n², a convergent p-series

No: it behaves like ∑ 1/√n, which diverges (limit comparison)
No: it behaves like ∑ 1/√n, which diverges (limit comparison)

Yes: it is smaller than the convergent ∑ 1/n²
Yes: it is smaller than the convergent ∑ 1/n²
Yes: it is smaller than the convergent ∑ 1/n²

4. Alternating series and the error bound

Yes, conditionally: the terms alternate, shrink and tend to 0

$\frac{5}{6}$
$\frac{35}{6}$

$\frac{1}{81}$
$\frac{1}{9}$

$\frac{1}{81}$
$\frac{1}{9}$
$\frac{1}{49}$

5. The ratio test; absolute and conditional convergence

$\frac{1}{8}$
$\frac{1}{4}$
$\frac{1}{5}$

Nothing: the ratio test is inconclusive when L = 1

Absolutely: ∑ 1/n² converges

$\frac{1}{8}$
$\frac{1}{4}$
$\frac{1}{5}$

6. Choosing a convergence test

The ratio test

The p-series test: p = 6 > 1, so it converges

The alternating series test
The alternating series test

The ratio test
The ratio test
The ratio test