The nth Term Test and the Integral Test

Most series have no formula for their partial sums, so their sums can't be found the way a geometric series's can. A convergence test answers a smaller question: whether the series converges at all.

The nth term test

\sum a_n

converges to

S

, then

S_n \to S

and

S_{n-1} \to S

. Each term is the difference of two partial sums,

a_n = S_n - S_{n-1}

, so

a_n \to S - S = 0

. Turned around, that's a test for divergence.

Example 1

Apply the nth term test to each series, and say what it shows.

(a)

\displaystyle\sum_{n=1}^{\infty} \Big(1 - \frac{3}{n}\Big)^n

(b)

\displaystyle\sum_{n=1}^{\infty} \frac{5^n}{4^n + 5^n}

(c)

\displaystyle\sum_{n=1}^{\infty} \frac{1}{\sqrt{2n + 7}}

(a) For

n \ge 4

the logarithm of the

n

th term is

n\ln\big(1 - \frac{3}{n}\big)

. Replace

n

by a real variable

x

, write it as a quotient, and apply L'Hospital's rule.

\begin{aligned} \lim_{x\to\infty} \frac{\ln(1 - 3/x)}{1/x} &= \lim_{x\to\infty} \frac{-3}{1 - 3/x} \\[4pt] &= -3 \end{aligned}

The terms follow the function's limit, so they approach

e^{-3} \ne 0

, and the series diverges.

(b) Divide the top and bottom by

5^n

\frac{5^n}{4^n + 5^n} = \frac{1}{(4/5)^n + 1} \to \frac{1}{0 + 1} = 1

The limit isn't 0, so the series diverges.

\lim_{n\to\infty} \frac{1}{\sqrt{2n + 7}} = 0

, the nth term test gives no conclusion about this series.

The integral test

When the terms come from a decreasing function, a series can be compared with an improper integral. Suppose

a_n = f(n)

for a decreasing function

f

. On

[n - 1, n]

the curve stays above

f(n)

, so a rectangle of height

a_n

and width 1 fits under it.

Rectangles of heights f(3), f(4), f(5), … under the curve from Example 2, starting at x = 2. The vertical scale is stretched. Their total area is at most the area under the curve from x = 2 on.

If the integral converges, the partial sums are increasing and stay below a fixed number, so they converge. Rectangles of height

a_n

[n, n + 1]

reach above the curve instead, so

a_N + \cdots + a_M \ge \int_N^{M+1} f(x)\,dx

. If the integral diverges, the partial sums grow without bound.

Dropping or changing finitely many terms never changes whether a series converges, so the conditions only need to hold from some

N

on. When the integral converges, its value is not the sum of the series.

Example 2

Use the integral test to decide whether

\sum_{n=1}^{\infty} \frac{n}{(n^2 + 4)^2}

converges.

Let

f(x) = \frac{x}{(x^2 + 4)^2}

. It's positive and continuous for

x > 0

. Check where it decreases.

f'(x) = \frac{4 - 3x^2}{(x^2 + 4)^3}

This is negative for

x > \frac{2}{\sqrt{3}}

, so

f

is decreasing for

x \ge 2

. Apply the test from

N = 2

\begin{aligned} &\int_{2}^{\infty} \frac{x}{(x^2 + 4)^2}\,dx \\[4pt] &\quad= \lim_{b\to\infty} \left[-\frac{1}{2(x^2 + 4)}\right]_{2}^{b} \\[4pt] &\quad= \lim_{b\to\infty} \left(\frac{1}{16} - \frac{1}{2(b^2 + 4)}\right) \\[4pt] &\quad= \frac{1}{16} \end{aligned}

The integral converges, so

\sum_{n=2}^{\infty} \frac{n}{(n^2 + 4)^2}

converges. Adding the term for

n = 1

doesn't change that, so the whole series converges. Its sum is about 0.119, while the integral is

\frac{1}{16}

Example 3

Decide whether

\sum_{n=1}^{\infty} \frac{1}{\sqrt{2n + 7}}

converges.

The terms approach 0, so the nth term test says nothing. Use the integral test. The function

f(x) = \frac{1}{\sqrt{2x + 7}}

is positive and continuous for

x \ge 1

, and decreasing because

2x + 7

increases.

\begin{aligned} &\int_{1}^{\infty} \frac{dx}{\sqrt{2x + 7}} \\[4pt] &\quad= \lim_{b\to\infty} \Big[\sqrt{2x + 7}\,\Big]_{1}^{b} \\[4pt] &\quad= \lim_{b\to\infty} \left(\sqrt{2b + 7} - 3\right) = \infty \end{aligned}

The integral diverges, so the series diverges too, even though its terms approach 0.