10.1 - Defining Convergent and Divergent Infinite Series

Sequences: The Fundamentals

A sequence is a function

a_n

that produces terms based on the position

n

, where

n

is a positive integer. The notation

a_n

means the

n

th term.

Example 1: Finding Terms

Given

a_n = 2n + (-1)^n \cdot 3

, find the first five terms.

a_1 = 2 - 3 = -1

a_2 = 4 + 3 = 7

a_3 = 6 - 3 = 3

a_4 = 8 + 3 = 11

a_5 = 10 - 3 = 7

The sequence is

-1, 7, 3, 11, 7, \ldots

One-to-one correspondence: a sequence puts its numbers in one-to-one correspondence with the positive integers. The 1st term is uniquely

a_1

, the 2nd is uniquely

a_2

, and so on. Every position has exactly one value.

Special Properties of Sequences

Monotonic Sequences

A sequence is monotonic if it satisfies one of these:

Monotonically increasing (non-decreasing):

a_1 \leq a_2 \leq a_3 \leq \cdots \leq a_n \leq a_{n+1} \leq \cdots

The terms never decrease. They can stay flat or go up.

Monotonically decreasing (non-increasing):

a_1 \geq a_2 \geq a_3 \geq \cdots \geq a_n \geq a_{n+1} \geq \cdots

The terms never increase.

10.1 - Defining Convergent and Divergent Infinite Series

Bounded Sequences

Bounded above: there’s an upper bound

M

with

a_n \leq M

for all

n

Bounded below: there’s a lower bound

m

with

a_n \geq m

for all

n

Bounded: both, so

m \leq a_n \leq M

for all

n

Infinite Series and Partial Sums

An infinite series is the sum of every term in an infinite sequence:

\sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + a_4 + \cdots

The $n$ th partial sum

S_n

is the sum of just the first

n

terms:

S_n = \sum_{k=1}^{n} a_k = a_1 + a_2 + \cdots + a_n

Critical Distinction: $a_n$ vs. $S_n$

$a_n$ is the $n$ th term, a single value

$S_n$ is the sum of the first $n$ terms, an accumulated total

Example 2

Given the sequence

2, 4, 6, 8, 10, \ldots

with

a_n = 2n

a_4 = 2(4) = 8

S_4 = 2 + 4 + 6 + 8 = 20

Series and Limits

\sum_{n=1}^{\infty} a_n = \lim_{n \to \infty} S_n

The sum of an infinite series is the limit of its partial sums.

Convergence and Divergence

For

\sum_{n=1}^{\infty} a_n

with partial sums

S_n

Convergent: the series converges to

L

\lim_{n \to \infty} S_n = L

where

L

is a finite number. We then write

\sum_{n=1}^{\infty} a_n = L

Divergent: the series diverges if

\lim_{n \to \infty} S_n

doesn’t exist or equals

\pm\infty

Checking by partial sums: compute several

S_n

and look for a pattern. If they seem to approach a specific number, the series likely converges. If they grow without bound, it diverges.

Example 3: Testing for Convergence

Does

\sum_{n=1}^{\infty} \frac{1}{3^n}

converge or diverge?

S_1 = \frac{1}{3} \approx 0.333

S_2 = \frac{4}{9} \approx 0.444

S_3 = \frac{13}{27} \approx 0.481

S_4 = \frac{40}{81} \approx 0.494

The partial sums look like they’re heading for

0.5

, which suggests convergence to $\frac{1}{2}$ .

Finding a Formula for $S_n$

For some series you can derive an explicit formula for

S_n

and then take the limit.

Example 4: Deriving $S_n$

Find

\sum_{n=1}^{\infty} \frac{1}{4^n}

Step 1: Write out $S_n$ .

S_n = \frac{1}{4} + \frac{1}{4^2} + \frac{1}{4^3} + \cdots + \frac{1}{4^n}

Step 2: Multiply by $\frac{1}{4}$ .

\frac{1}{4}S_n = \frac{1}{4^2} + \frac{1}{4^3} + \cdots + \frac{1}{4^{n+1}}

Step 3: Subtract. Almost everything cancels, leaving only the first term of

S_n

and the last term of

\frac{1}{4}S_n

S_n - \frac{1}{4}S_n = \frac{1}{4} - \frac{1}{4^{n+1}}

Step 4: Simplify.

\frac{3}{4}S_n = \frac{1}{4} - \frac{1}{4^{n+1}} \quad \Rightarrow \quad S_n = \frac{1}{3} - \frac{1}{3 \cdot 4^n}

Step 5: Take the limit. As

n \to \infty

\frac{1}{4^n} \to 0

, so

\lim_{n \to \infty} S_n = \frac{1}{3}

The series converges to

\frac{1}{3}

Using Technology for Partial Sums

For series without a nice closed form, approximate big partial sums on the calculator.

Example 5: Calculator Method

Find

S_{150}

for

\sum_{n=1}^{\infty} \frac{8}{n^2 + 3n}

TI-84 steps:

Press MATH, scroll to 0: sum(

Enter sum(seq(8/(X²+3X), X, 1, 150))

Press ENTER

Result:

S_{150} \approx 4.836

Try larger values:

$S_{500} \approx 4.873$

$S_{1000} \approx 4.881$

The partial sums are creeping toward about

4.889

, which suggests convergence. (This one actually telescopes to exactly

\frac{44}{9} \approx 4.889

Example 6: A Divergent Series

Does

\sum_{n=1}^{\infty} n

converge or diverge?

S_1 = 1, \quad S_2 = 3, \quad S_3 = 6, \quad S_4 = 10, \quad S_5 = 15

The partial sums keep climbing with no ceiling. They’re monotonically increasing but not bounded above.

\lim_{n \to \infty} S_n = \infty

The series diverges.

Example 7: Another Convergent Series

Does

\sum_{n=1}^{\infty} \frac{3}{n(n+1)}

converge or diverge?

S_1 = \frac{3}{2} = 1.5

S_2 = 1.5 + 0.5 = 2.0

S_3 = 2.0 + 0.25 = 2.25

S_4 = 2.25 + 0.15 = 2.40

S_5 = 2.40 + 0.10 = 2.50

The partial sums

1.5, 2.0, 2.25, 2.40, 2.50, \ldots

appear to approach

3

The series converges to 3.

To be sure, split each term:

\frac{3}{n(n+1)} = \frac{3}{n} - \frac{3}{n+1}

. Then the partial sums telescope (almost everything cancels):

S_n = \left(3 - \frac{3}{2}\right) + \left(\frac{3}{2} - \frac{3}{3}\right) + \cdots + \left(\frac{3}{n} - \frac{3}{n+1}\right) = 3 - \frac{3}{n+1}

Since

\lim_{n \to \infty} \left(3 - \frac{3}{n+1}\right) = 3

, the limit of the partial sums exists and equals 3, so the series converges to exactly 3. (Check:

S_5 = 3 - \frac{3}{6} = 2.5

, matching the list above.)

Summary of Strategies

1. Calculate several partial sums. Look for a pattern. Approaching a number suggests convergence; growing without bound means divergence.

2. Find a formula for $S_n$ if you can. Use the multiply-and-subtract trick, then take

\lim_{n\to\infty} S_n

. A finite limit means convergence; an infinite one or none at all means divergence.

3. Use technology for big partial sums. Compute

S_{100}

S_{500}

S_{1000}

and see whether the values stabilize.

Key Takeaways

$a_n$ is the $n$ th term (a single value); $S_n$ is the sum of the first $n$ terms

An infinite series $\sum_{n=1}^{\infty} a_n$ is the sum of infinitely many terms

Convergent means $\lim_{n \to \infty} S_n = L$ for a finite $L$

Divergent means that limit doesn’t exist or is $\pm\infty$

$\sum_{n=1}^{\infty} a_n = \lim_{n \to \infty} S_n$

Monotonic sequences never increase, or never decrease

Bounded sequences have both upper and lower bounds

Test convergence by computing partial sums and watching the pattern

The multiply-and-subtract technique can give you a formula for $S_n$

Technology helps when there’s no closed form

Back Next

10.1 - Defining Convergent and Divergent Infinite Series

Sequences: The Fundamentals

Example 1: Finding Terms

Special Properties of Sequences

Monotonic Sequences

Bounded Sequences

Infinite Series and Partial Sums

Critical Distinction: ana_nan​ vs. SnS_nSn​

Example 2

Series and Limits

Convergence and Divergence

Example 3: Testing for Convergence

Finding a Formula for SnS_nSn​

Example 4: Deriving SnS_nSn​

Using Technology for Partial Sums

Example 5: Calculator Method

Example 6: A Divergent Series

Example 7: Another Convergent Series

Summary of Strategies

Key Takeaways

Critical Distinction: $a_n$ vs. $S_n$

Finding a Formula for $S_n$

Example 4: Deriving $S_n$