The Integral Test
Solved Problems
Example 3.
Determine whether the series \[\sum\limits_{n = 1}^\infty {\frac{1}{{\left( {n + 1} \right)\ln \left( {n + 1} \right)}}}\] converges or diverges.
Solution.
We use the integral test. Calculate the improper integral
\[
\int\limits_1^\infty {\frac{{dx}}{{\left( {x + 1} \right)\ln \left( {x + 1} \right)}}}
= \lim\limits_{n \to \infty } \int\limits_1^n {\frac{{dx}}{{\left( {x + 1} \right)\ln \left( {x + 1} \right)}}}
= \lim\limits_{n \to \infty } \int\limits_1^n {\frac{{d\left( {x + 1} \right)}}{{\left( {x + 1} \right)\ln \left( {x + 1} \right)}}}
= \lim\limits_{n \to \infty } \int\limits_1^n {\frac{{d\ln \left( {x + 1} \right)}}{{\ln \left( {x + 1} \right)}}}
= \lim\limits_{n \to \infty } \left. {\left[ {\ln \ln \left( {x + 1} \right)} \right]} \right|_1^n
= \lim\limits_{n \to \infty } \left[ {\ln \ln \left( {n + 1} \right) - \ln \ln 2} \right] = \infty .\]
The integral approaches infinity. Therefore, the initial series diverges.
Example 4.
Investigate the series \[\sum\limits_{n = 1}^\infty {\frac{n}{{{n^2} + 1}}} \] for convergence.
Solution.
We evaluate the improper integral:
\[\int\limits_1^\infty {\frac{{xdx}}{{{x^2} + 1}}} = \lim\limits_{n \to \infty } \int\limits_1^n {\frac{{xdx}}{{{x^2} + 1}}} .\]
Make the substitution: \(t = {x^2} + 1.\) Then
\[dt = 2xdx,\;\;\Rightarrow xdx =\frac{{dt}}{2}.\]
The integral becomes
\[
\lim\limits_{n \to \infty } \int\limits_1^n {\frac{{xdx}}{{{x^2} + 1}}}
= \lim\limits_{n \to \infty } \int\limits_2^{{n^2} + 1} {\frac{{\frac{{dt}}{2}}}{t}}
= \frac{1}{2}\lim\limits_{n \to \infty } \int\limits_2^{{n^2} + 1} {\frac{{dt}}{t}}
= \frac{1}{2}\lim\limits_{n \to \infty } \left. {\left( {\ln t} \right)} \right|_2^{{n^2} + 1}
= \frac{1}{2}\lim\limits_{n \to \infty } \left[ {\ln \left( {{n^2} + 1} \right) - \ln 2} \right] = \infty .\]
Since the given integral diverges, the initial series \(\sum\limits_{n = 1}^\infty {{\frac{n}{{{n^2} + 1}}}} \) also diverges by the integral test.
Example 5.
Determine whether \[\sum\limits_{n = 1}^\infty {\frac{{\arctan n}}{{1 + {n^2}}}}\] converges or diverges.
Solution.
We easily can see that \(\arctan n \lt {\frac{\pi }{2}}.\) Hence, by comparison test,
\[\sum\limits_{n = 1}^\infty {\frac{{\arctan n}}{{1 + {n^2}}}} \lt \sum\limits_{n = 1}^\infty {\frac{{\frac{\pi }{2}}}{{1 + {n^2}}}} = \frac{\pi }{2}\sum\limits_{n = 1}^\infty {\frac{1}{{1 + {n^2}}}} .\]
Use the integral test to determine convergence of the series \(\sum\limits_{n = 1}^\infty {\frac{1}{{1 + {n^2}}}}:\)
\[
\int\limits_1^\infty {\frac{{dx}}{{1 + {x^2}}}}
= \lim\limits_{n \to \infty } \int\limits_1^n {\frac{{dx}}{{1 + {x^2}}}}
= \lim\limits_{n \to \infty } \left. {\left( {\arctan x} \right)} \right|_1^n
= \lim\limits_{n \to \infty } \left( {\arctan n - \arctan 1} \right)
= \frac{\pi }{2} - \frac{\pi }{4} = \frac{\pi }{4}.\]
Since the improper integral is convergent, then the original series is also convergent.
Example 6.
Investigate whether the series \[\sum\limits_{n = 0}^\infty {n{e^{ - n}}}\] converges or diverges.
Solution.
Using the integral test, calculate the improper integral
\[\int\limits_0^\infty {x{e^{ - x}}dx} = \lim\limits_{n \to \infty } \int\limits_0^n {x{e^{ - x}}dx} .\]
Apply integration by parts:
\[u = x,\;\;dv = {e^{ - x}}dx,\;\; \Rightarrow du = dx,\;\; v = \int {{e^{ - x}}dx} = - {e^{ - x}}.\]
Then we obtain
\[
\lim\limits_{n \to \infty } \int\limits_0^n {x{e^{ - x}}dx}
= \lim\limits_{n \to \infty } \left[ {\left. {\left( { - x{e^{ - x}}} \right)} \right|_0^n - \int\limits_0^n {\left( { - {e^{ - x}}} \right)dx} } \right]
= \lim\limits_{n \to \infty } \left[ {\left. {\left( { - x{e^{ - x}}} \right)} \right|_0^n + \int\limits_0^n {{e^{ - x}}dx} } \right]
= \lim\limits_{n \to \infty } \left. {\left( { - x{e^{ - x}} - {e^{ - x}}} \right)} \right|_0^n
= - \lim\limits_{n \to \infty } \left. {\left[ {{e^{ - x}}\left( {x + 1} \right)} \right]} \right|_0^n
= - \lim\limits_{n \to \infty } \left[ {{e^{ - n}}\left( {n + 1} \right) - 1} \right]
= 1 - \lim\limits_{n \to \infty } \frac{{n + 1}}{{{e^n}}}.\]
By L'Hopital's rule, the limit in the last expression is
\[\lim\limits_{n \to \infty } \frac{{n + 1}}{{{e^n}}} \sim \lim\limits_{x \to \infty } \frac{{x + 1}}{{{e^x}}} = \lim\limits_{x \to \infty } \frac{{{{\left( {x + 1} \right)}^\prime }}}{{{{\left( {{e^x}} \right)}^\prime }}} = \lim\limits_{x \to \infty } \frac{1}{{{e^x}}} = 0.\]
Hence, the improper integral is finite and equal to \(1.\) Therefore, the original series is convergent.
