Defining Convergent and Divergent Infinite Series

AP Calculus BC · Infinite Sequences and Series

Defining Convergent and Divergent Infinite Series

A series is the limit of its partial sums — nothing is ever added infinitely many times.

key idea$\sum_{n=1}^{\infty}a_n=\lim_{N\to\infty}S_N,\qquad S_N=\sum_{n=1}^{N}a_n$

What it says

Partial sum

\[S_N=a_1+a_2+\cdots+a_N\]

A finite sum: add the first $N$ terms and stop. Every partial sum is an ordinary number, and the sequence $S_1,S_2,S_3,\dots$ is what the whole subject is really about.

Convergent series

\[\sum_{n=1}^{\infty}a_n=L\iff \lim_{N\to\infty}S_N=L\]

An infinite series converges when its sequence of partial sums approaches a finite limit. The sum of a series is therefore defined as a limit — nothing is ever added infinitely many times.

SequenceAn ordered list of numbers $a_1,a_2,\dots$ — no addition involved.

SeriesThe sum of a sequence's terms. A series converges or diverges; a sequence converges or diverges. They are different objects with different answers.

Partial sum $S_N$The finite sum of the first $N$ terms. The sequence of partial sums is what is tested for a limit.

ConvergesThe partial sums approach a finite value $L$, and $L$ is called the sum.

DivergesThe partial sums fail to approach any finite value — they may grow without bound or oscillate.

Telescoping seriesOne whose partial sums collapse because interior terms cancel, leaving a short expression whose limit can be read directly.

Series	Terms $a_n\to$	Partial sums	Verdict
$\displaystyle\sum\frac{1}{2^{n}}$	$0$	$\to1$	converges to $1$
$\displaystyle\sum\frac{1}{n}$	$0$	grow without bound	diverges
$\displaystyle\sum(-1)^{n}$	does not exist	oscillate $-1,0,-1,\dots$	diverges
$\displaystyle\sum n$	$\infty$	$\to\infty$	diverges
$\displaystyle\sum\left(\frac1n-\frac1{n+1}\right)$	$0$	$1-\frac{1}{N+1}\to1$	converges to $1$

Rows 1 and 2 are the pair worth staring at. Both have terms going to zero, and only one converges — which is why ‘the terms shrink’ can never be the end of an argument.

what the definition buys you

A series is the limit of its partial sums. That definition answers every conceptual question in this unit.
$a_n\to0$ is necessary but not sufficient for convergence.
A sequence and its series can behave differently: $a_n=\dfrac1n$ converges to $0$ while $\sum\dfrac1n$ diverges.
Telescoping and geometric are the only two families whose exact sums you are expected to produce.

Reading the notation

$a_n$one term

$S_N$a finite sum of the first $N$

$\sum_{n=1}^{\infty}a_n$the limit of the $S_N$

Worked examples

EX 1

Convergence from the partial sums

a telescoping series

Determine whether $\displaystyle\sum_{n=1}^{\infty}\left(\dfrac1n-\dfrac{1}{n+1}\right)$ converges, and find its sum.

Write out a partial sum

$S_N=\left(1-\tfrac12\right)+\left(\tfrac12-\tfrac13\right)+\cdots+\left(\tfrac1N-\tfrac1{N+1}\right)$.

Cancel the interior

Every negative term is killed by the next positive one, leaving $S_N=1-\dfrac{1}{N+1}$.

Take the limit

$\displaystyle\lim_{N\to\infty}\left(1-\dfrac{1}{N+1}\right)=1$.

State the conclusion

The series converges, with sum $1$.

Note the method

The exact sum was available only because the partial sums collapsed to a short formula. That is rare, and it is the whole reason telescoping is worth spotting.

EX 2

Terms to zero, series diverges

the harmonic series

Explain why $\displaystyle\sum_{n=1}^{\infty}\dfrac1n$ diverges even though $\dfrac1n\to0$.

Group the terms

$1+\tfrac12+\left(\tfrac13+\tfrac14\right)+\left(\tfrac15+\cdots+\tfrac18\right)+\cdots$

Bound each group below

$\tfrac13+\tfrac14\gt\tfrac14+\tfrac14=\tfrac12$, and the next group also exceeds $\tfrac12$, and so on.

Add up the bounds

Each block contributes more than $\tfrac12$, and there are infinitely many blocks.

Conclude

$S_N$ grows without bound, so the series diverges.

Draw the lesson

The terms shrink, but not fast enough. Convergence is about the rate at which terms shrink, never about the fact that they do.

EX 3

Oscillating partial sums

no limit at all

Determine whether $\displaystyle\sum_{n=1}^{\infty}(-1)^{n}$ converges.

List the partial sums

$S_1=-1$, $S_2=0$, $S_3=-1$, $S_4=0,\dots$

Look for a limit

The sequence alternates between $-1$ and $0$ forever.

Apply the definition

A limit must be a single value. This sequence has none.

Conclude

The series diverges.

Note the second reason

The terms themselves do not approach zero, which is already enough to force divergence — two independent arguments for the same verdict.

EX 4

Sequence versus series

two different questions

For $a_n=\dfrac{n}{n+1}$, discuss the convergence of the sequence and of the series.

Test the sequence

$\displaystyle\lim_{n\to\infty}\dfrac{n}{n+1}=1$, so the sequence converges to $1$.

Test the series

The terms approach $1$, not $0$.

Apply the necessary condition

For a series to converge its terms must go to zero. These do not.

Conclude

The series diverges, even though the sequence converges.

Fix the distinction

‘Does $a_n$ converge?’ and ‘does $\sum a_n$ converge?’ are separate questions with separate answers. Read which one is being asked.

Boundary cases

Terms going to zero proves nothing

$\sum\dfrac1n$ has terms tending to $0$ and still diverges. The condition is necessary, not sufficient.

The sum is a limit, not an infinite addition

Every partial sum is finite. Convergence means those finite numbers settle down, which is what makes the notion well defined.

A sequence and its series are different objects

$a_n\to L$ describes the terms; $\sum a_n$ describes their running total. Only $L=0$ leaves any hope for the series.

Exact sums are rare

Outside geometric and telescoping series, expect to determine whether a series converges, not what it converges to.

Course map. Topic 10.1 · LO LIM-7.A — determine whether a series converges or diverges. BC-only.