The Integral Test

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Let \(f\left( x \right)\) be a function which is continuous, positive, and decreasing for all \(x\) in the range \(\left[ {1, + \infty } \right).\) Then the series

\[\sum\limits_{n = 1}^\infty {f\left( n \right)} = f\left( 1 \right) + f\left( 2 \right) + f\left( 3 \right) + \ldots + f\left( n \right) + \ldots \]

converges if the improper integral \(\int\limits_1^\infty {f\left( x \right)dx}\) converges, and diverges if \(\int\limits_1^\infty {f\left( x \right)dx} \to \infty.\)

Solved Problems

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Example 1

Determine whether the series \[\sum\limits_{n = 1}^\infty {\frac{1}{{1 + 10n}}}\] converges or diverges.

Example 2

Show that the \(p\)-series \[\sum\limits_{n = 1}^\infty {\frac{1}{{{n^p}}}} \] converges for \(p \gt 1.\)

Example 1.

Determine whether the series \[\sum\limits_{n = 1}^\infty {\frac{1}{{1 + 10n}}}\] converges or diverges.

Solution.

We use the integral test. Calculate the improper integral

\[ \int\limits_1^\infty {\frac{{dx}}{{1 + 10x}}} = \lim\limits_{n \to \infty } \int\limits_1^n {\frac{{dx}}{{1 + 10x}}} = \lim\limits_{n \to \infty } \left. {\left[ {\frac{1}{{10}}\ln \left( {1 + 10x} \right)} \right]} \right|_1^n = \frac{1}{{10}}\lim\limits_{n \to \infty } \left[ {\ln \left( {1 + 10n} \right) - \ln 11} \right] = \infty .\]

Thus, the given series is divergent.

Example 2.

Show that the \(p\)-series \[\sum\limits_{n = 1}^\infty {\frac{1}{{{n^p}}}} \] converges for \(p \gt 1.\)

Solution.

We consider the corresponding function \(f\left( x \right) = \frac{1}{{{x^p}}}\) and apply the integral test. The improper integral is

\[ \int\limits_1^\infty {\frac{{dx}}{{{x^p}}}} = \lim\limits_{n \to \infty } \int\limits_1^n {\frac{{dx}}{{{x^p}}}} = \lim\limits_{n \to \infty } \int\limits_1^n {{x^{ - p}}dx} = \lim\limits_{n \to \infty } \left. {\left( {\frac{1}{{ - p + 1}}{x^{ - p + 1}}} \right)} \right|_1^n = \frac{1}{{1 - p}}\lim\limits_{n \to \infty } \left. {\left( {\frac{1}{{{x^{p - 1}}}}} \right)} \right|_1^n = \frac{1}{{1 - p}}\lim\limits_{n \to \infty } \left( {\frac{1}{{{n^{p - 1}}}} - 1} \right) = \frac{1}{{p - 1}}.\]

Hence, the \(p\)-series converges for \(p \gt 1.\)