5.3 The divergence and integral tests (Page 3/9)

Calculus volume 2 Page 3 / 9

Although convergence of $\int_{N}^{\infty} f (x) d x$ implies convergence of the related series $\sum_{n = 1}^{\infty} a_{n},$ it does not imply that the value of the integral and the series are the same. They may be different, and often are. For example,

\sum_{n = 1}^{\infty} {(\frac{1}{e})}^{n} = \frac{1}{e} + {(\frac{1}{e})}^{2} + {(\frac{1}{e})}^{3} + ⋯

is a geometric series with initial term $a = 1 / e$ and ratio $r = 1 / e,$ which converges to

\frac{1 / e}{1 - (1 / e)} = \frac{1 / e}{(e - 1) / e} = \frac{1}{e - 1} .

However, the related integral $\int_{1}^{\infty} {(1 / e)}^{x} d x$ satisfies

\int_{1}^{\infty} {(\frac{1}{e})}^{x} d x = \int_{1}^{\infty} e^{− x} d x = lim_{b \to \infty} \int_{1}^{b} e^{− x} d x = lim_{b \to \infty} - e^{− x} |_{1}^{b} = lim_{b \to \infty} [− e^{− b} + e^{−1}] = \frac{1}{e} .

Using the integral test

For each of the following series, use the integral test to determine whether the series converges or diverges.

$\sum_{n = 1}^{\infty} 1 / n^{3}$
$\sum_{n = 1}^{\infty} 1 / \sqrt{2 n - 1}$

Compare

$\sum_{n = 1}^{\infty} \frac{1}{n^{3}} and \int_{1}^{\infty} \frac{1}{x^{3}} d x .$

We have

$\int_{1}^{\infty} \frac{1}{x^{3}} d x = lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^{3}} d x = lim_{b \to \infty} [- \frac{1}{2 x^{2}} |_{1}^{b}] = lim_{b \to \infty} [- \frac{1}{2 b^{2}} + \frac{1}{2}] = \frac{1}{2} .$

Thus the integral $\int_{1}^{\infty} 1 / x^{3} d x$ converges, and therefore so does the series

$\sum_{n = 1}^{\infty} \frac{1}{n^{3}} .$
Compare

$\sum_{n = 1}^{\infty} \frac{1}{\sqrt{2 n - 1}} and \int_{1}^{\infty} \frac{1}{\sqrt{2 x - 1}} d x .$

Since

$\begin{matrix} \int_{1}^{\infty} \frac{1}{\sqrt{2 x - 1}} d x & = lim_{b \to \infty} \int_{1}^{b} \frac{1}{\sqrt{2 x - 1}} d x = lim_{b \to \infty} \sqrt{2 x - 1} |_{1}^{b} \\ = lim_{b \to \infty} [\sqrt{2 b - 1} - 1] = \infty, \end{matrix}$

the integral $\int_{1}^{\infty} 1 / \sqrt{2 x - 1} d x$ diverges, and therefore

$\sum_{n = 1}^{\infty} \frac{1}{\sqrt{2 n - 1}}$

diverges.

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Use the integral test to determine whether the series $\sum_{n = 1}^{\infty} \frac{n}{3 n^{2} + 1}$ converges or diverges.

The series diverges.

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The p -series

The harmonic series $\sum_{n = 1}^{\infty} 1 / n$ and the series $\sum_{n = 1}^{\infty} 1 / n^{2}$ are both examples of a type of series called a p -series.

Definition

For any real number $p,$ the series

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

is called a p -series .

We know the p -series converges if $p = 2$ and diverges if $p = 1 .$ What about other values of $p ?$ In general, it is difficult, if not impossible, to compute the exact value of most $p$ -series. However, we can use the tests presented thus far to prove whether a $p$ -series converges or diverges.

If $p < 0,$ then $1 / n^{p} \to \infty,$ and if $p = 0,$ then $1 / n^{p} \to 1 .$ Therefore, by the divergence test,

\sum_{n = 1}^{\infty} 1 / n^{p} diverges if p \leq 0 .

If $p > 0,$ then $f (x) = 1 / x^{p}$ is a positive, continuous, decreasing function. Therefore, for $p > 0,$ we use the integral test, comparing

\sum_{n = 1}^{\infty} \frac{1}{n^{p}} and \int_{1}^{\infty} \frac{1}{x^{p}} d x .

We have already considered the case when $p = 1 .$ Here we consider the case when $p > 0, p \neq 1 .$ For this case,

\int_{1}^{\infty} \frac{1}{x^{p}} d x = lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^{p}} d x = lim_{b \to \infty} \frac{1}{1 - p} x^{1 - p} |_{1}^{b} = lim_{b \to \infty} \frac{1}{1 - p} [b^{1 - p} - 1] .

Because

b^{1 - p} \to 0 if p > 1 and b^{1 - p} \to \infty if p < 1,

we conclude that

\int_{1}^{\infty} \frac{1}{x^{p}} d x = {\begin{matrix} \frac{1}{p - 1} if p > 1 \\ \infty if p < 1 \end{matrix} .

Therefore, $\sum_{n = 1}^{\infty} 1 / n^{p}$ converges if $p > 1$ and diverges if $0 < p < 1 .$

In summary,

\sum_{n = 1}^{\infty} \frac{1}{n^{p}} {\begin{cases} converges if p > 1 \\ diverges if p \leq 1 \end{cases} .

Testing for convergence of p -series

For each of the following series, determine whether it converges or diverges.

$\sum_{n = 1}^{\infty} \frac{1}{n^{4}}$
$\sum_{n = 1}^{\infty} \frac{1}{n^{2 / 3}}$

This is a p -series with $p = 4 > 1,$ so the series converges.
Since $p = 2 / 3 < 1,$ the series diverges.

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Does the series $\sum_{n = 1}^{\infty} \frac{1}{n^{5 / 4}}$ converge or diverge?

The series converges.

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Estimating the value of a series

Suppose we know that a series $\sum_{n = 1}^{\infty} a_{n}$ converges and we want to estimate the sum of that series. Certainly we can approximate that sum using any finite sum $\sum_{n = 1}^{N} a_{n}$ where $N$ is any positive integer. The question we address here is, for a convergent series $\sum_{n = 1}^{\infty} a_{n},$ how good is the approximation $\sum_{n = 1}^{N} a_{n} ?$ More specifically, if we let

R_{N} = \sum_{n = 1}^{\infty} a_{n} - \sum_{n = 1}^{N} a_{n}

be the remainder when the sum of an infinite series is approximated by the $N th$ partial sum, how large is $R_{N} ?$ For some types of series, we are able to use the ideas from the integral test to estimate $R_{N} .$

Remainder estimate from the integral test

Suppose $\sum_{n = 1}^{\infty} a_{n}$ is a convergent series with positive terms. Suppose there exists a function $f$ satisfying the following three conditions:

$f$ is continuous,
$f$ is decreasing, and
$f (n) = a_{n}$ for all integers $n \geq 1 .$

Let $S_{N}$ be the N th partial sum of $\sum_{n = 1}^{\infty} a_{n} .$ For all positive integers $N,$

S_{N} + \int_{N + 1}^{\infty} f (x) d x < \sum_{n = 1}^{\infty} a_{n} < S_{N} + \int_{N}^{\infty} f (x) d x .

In other words, the remainder $R_{N} = \sum_{n = 1}^{\infty} a_{n} - S_{N} = \sum_{n = N + 1}^{\infty} a_{n}$ satisfies the following estimate:

\int_{N + 1}^{\infty} f (x) d x < R_{N} < \int_{N}^{\infty} f (x) d x .

This is known as the remainder estimate .

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Source: OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2

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