Math Core

What does it mean to add infinitely many numbers? You can never finish the addition, but you can add more and more terms and watch the running total. If those running totals settle down to a number, that number is the sum. This lesson makes that idea precise and gives you the first, quickest test for divergence.

Series and partial sums

An infinite series is an expression of the form

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

To give it meaning, build the sequence of partial sums:

S_1 = a_1, \qquad S_2 = a_1 + a_2, \qquad S_3 = a_1 + a_2 + a_3, \qquad \dots, \qquad S_n = \sum_{k=1}^{n} a_k.

Definition

Convergent series

The series $\displaystyle\sum_{n=1}^{\infty} a_n$ converges to the sum $S$ if its sequence of partial sums converges to $S$ :

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

If the partial sums do not approach a finite number, the series diverges.

So every series question is really a sequence question, just about the sequence $S_n$ instead of the sequence $a_n$ . Keep the two sequences separate in your mind: $a_n$ is the list of things you add, and $S_n$ is the list of running totals.

A first example: telescoping

Consider $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ . Using partial fractions, $\dfrac{1}{n(n+1)} = \dfrac{1}{n} - \dfrac{1}{n+1}$ , so

S_n = \left(1 - \tfrac{1}{2}\right) + \left(\tfrac{1}{2} - \tfrac{1}{3}\right) + \left(\tfrac{1}{3} - \tfrac{1}{4}\right) + \cdots + \left(\tfrac{1}{n} - \tfrac{1}{n+1}\right).

Almost everything cancels, leaving $S_n = 1 - \dfrac{1}{n+1}$ . As $n \to \infty$ , $S_n \to 1$ , so the series converges and its sum is $1$ . A series whose partial sums collapse like this is called telescoping.

Partial sums S_n of the telescoping series Σ 1/(n(n+1)) approach the sum 1.Open in grapher →

The nth term test for divergence

Suppose a series converges, with $S_n \to S$ . Each term is the difference of two consecutive partial sums, $a_n = S_n - S_{n-1}$ , and both of those approach $S$ . So $a_n \to S - S = 0$ . In words: if you are adding up infinitely many numbers and getting a finite total, the numbers you add must be shrinking to zero.

Turn that around and you get a divergence test.

nth term test for divergence

If $\displaystyle\lim_{n\to\infty} a_n \ne 0$ (or the limit does not exist), then $\displaystyle\sum a_n$ diverges.

If $\displaystyle\lim_{n\to\infty} a_n = 0$ , the test tells you nothing. The series might converge or might diverge.

Common mistake

The nth term test can only prove divergence. It can never prove convergence. The classic example is the harmonic series $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}$ : its terms go to $0$ , yet it diverges (you will prove this with the integral test soon). On the AP exam, writing " $a_n \to 0$ , so the series converges" is always marked wrong.

Algebra of series

Convergent series can be combined term by term, just like finite sums.

If $\sum a_n = A$ and $\sum b_n = B$ both converge, and $c$ is a constant, then

\sum (a_n + b_n) = A + B, \qquad \sum (a_n - b_n) = A - B, \qquad \sum c\,a_n = cA.

Two more facts are used constantly:

Adding or removing finitely many terms does not change whether a series converges (though it does change the sum). So $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=50}^{\infty} a_n$ either both converge or both diverge.
If $\sum a_n$ converges and $\sum b_n$ diverges, then $\sum (a_n + b_n)$ diverges.

Worked examples

Worked example: Applying the nth term test

Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{4n^2 + 1}{3n^2 - n}$ converges or diverges.

Solution. The terms approach

\lim_{n\to\infty} \frac{4n^2 + 1}{3n^2 - n} = \frac{4}{3} \ne 0.

By the nth term test, the series diverges. (You are adding numbers close to $\frac{4}{3}$ infinitely often, so of course the total grows without bound.)

Worked example: Partial sums from a formula

The $n$ th partial sum of a series $\sum_{n=1}^{\infty} a_n$ is $S_n = \dfrac{5n}{2n + 3}$ . Find the sum of the series and the value of $a_4$ .

Solution. The sum is the limit of the partial sums:

\sum_{n=1}^{\infty} a_n = \lim_{n\to\infty} \frac{5n}{2n+3} = \frac{5}{2}.

For the individual term, $a_4 = S_4 - S_3 = \dfrac{20}{11} - \dfrac{15}{9} = \dfrac{20}{11} - \dfrac{5}{3} = \dfrac{60 - 55}{33} = \dfrac{5}{33}$ .

Worked example: A telescoping sum

Find the sum of $\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}\right)$ .

Solution. Write out the partial sum:

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

As $n \to \infty$ , $S_n \to 1$ . The series converges to $1$ .

Tip

For a telescoping series, always write out the first three or four terms and the last one or two. Seeing which pieces survive is much safer than guessing.