Calculus

Infinite Sequences and Series

Sequences and Series Logo

Alternating Series

Solved Problems

Example 3.

Determine whether \[\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{n!}}} \] is absolutely convergent, conditionally convergent, or divergent.

Solution.

Applying the ratio test to the series with the corresponding nonnegative terms, we have

\[\lim\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \lim\limits_{n \to \infty } \frac{{\frac{1}{{\left( {n + 1} \right)!}}}}{{\frac{1}{{n!}}}} = \lim\limits_{n \to \infty } \frac{{n!}}{{\left( {n + 1} \right)!}} = \lim\limits_{n \to \infty } \frac{1}{{n + 1}} = 0.\]

Hence, the series is absolutely convergent.

Example 4.

Determine whether the alternating series \[\sum\limits_{n = 2}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}\sqrt n }}{{\ln n}}} \] is absolutely convergent, conditionally convergent, or divergent.

Solution.

First using the alternating series test, we find the limit \(\lim\limits_{n \to \infty } \left| {{a_n}} \right| \) \(= \lim\limits_{n \to \infty } \frac{{\sqrt n }}{{\ln n}} .\) Calculate this limit by L'Hopital's rule:

\[\lim\limits_{n \to \infty } \frac{{\sqrt n }}{{\ln n}} \sim \lim\limits_{x \to \infty } \frac{{\sqrt x }}{{\ln x}} = \lim\limits_{x \to \infty } \frac{{\frac{1}{{2\sqrt x }}}}{{\frac{1}{x}}} = \frac{1}{2}\lim\limits_{x \to \infty } \frac{x}{{\sqrt x }} = \frac{1}{2}\lim\limits_{x \to \infty } \sqrt x = \infty .\]

Thus, the given series is divergent.

Example 5.

Determine the \(n\)th term and test for convergence the series \[\frac{2}{{3!}} - \frac{{{2^2}}}{{5!}} + \frac{{{2^3}}}{{7!}} - \frac{{{2^4}}}{{9!}} + \ldots\]

Solution.

The \(n\)th term of the series is

\[{a_n} = {\left( { - 1} \right)}^{n + 1}\frac{{{2^n}}}{{\left( {2n + 1} \right)!}}.\]

Apply the ratio test to the series \(\sum\limits_{n = 1}^\infty {\left| {{a_n}} \right|} \) \(= \sum\limits_{n = 1}^\infty {\frac{{{2^n}}}{{\left( {2n + 1} \right)!}}}\) with nonnegative terms:

\[\lim\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \lim\limits_{n \to \infty } \frac{{\frac{{{2^{n + 1}}}}{{\left( {2n + 3} \right)!}}}}{{\frac{{{2^n}}}{{\left( {2n + 1} \right)!}}}} = \lim\limits_{n \to \infty } \frac{{{2^{n + 1}}\left( {2n + 1} \right)!}}{{{2^n}\left( {2n + 3} \right)!}} = \lim\limits_{n \to \infty } \frac{2}{{\left( {2n + 2} \right)\left( {2n + 3} \right)}} = 0.\]

Hence, the given series is absolutely convergent.

Example 6.

Investigate whether the series \[\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{5n - 1}} } \] is absolutely convergent, conditionally convergent, or divergent.

Solution.

Using the alternating series test, we see that the series is convergent:

\[\lim\limits_{n \to \infty } \left| {{a_n}} \right| = \lim\limits_{n \to \infty } \left| {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{5n - 1}}} \right| = \lim\limits_{n \to \infty } \frac{1}{{5n - 1}} = 0.\]

Now consider convergence of the series \(\sum\limits_{n = 1}^\infty {\left| {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{5n - 1}}} \right|} \) \(= \sum\limits_{n = 1}^\infty {\frac{1}{{5n - 1}}} \) with nonnegative terms. By the integral test, we have

\[ \int\limits_1^\infty {\frac{{dx}}{{5x - 1}}} = \lim\limits_{n \to \infty } \int\limits_1^n {\frac{{dx}}{{5x - 1}}} = \lim\limits_{n \to \infty } \left[ {\left. {\frac{1}{5}\left( {\ln \left| {5x - 1} \right|} \right)} \right|_1^n} \right] = \frac{1}{5}\lim\limits_{n \to \infty } \left[ {\ln \left( {5n - 1} \right) - \ln 4} \right] = \infty .\]

Hence, the series\(\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{5n - 1}}} \) is conditionally convergent.

Example 7.

Determine whether the alternating series \[\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{\sqrt {n\left( {n + 1} \right)} }}} \] is absolutely convergent, conditionally convergent, or divergent.

Solution.

First we apply the alternating series test:

\[\lim\limits_{n \to \infty } \left| {{a_n}} \right| = \lim\limits_{n \to \infty } \frac{1}{{\sqrt {n\left( {n + 1} \right)} }} = 0.\]

Hence, the given series is convergent. Now we determine is it absolutely or conditionally convergent. We will use the limit comparison test and compare the corresponding series with nonnegative terms \(\sum\limits_{n = 1}^\infty {\left| {{a_n}} \right|} \) \(= \sum\limits_{n = 1}^\infty {\frac{1}{{\sqrt {n\left( {n + 1} \right)} }}} \) with the divergent harmonic series \(\sum\limits_{n = 1}^\infty {\frac{1}{n}}:\)

\[\lim\limits_{n \to \infty } \frac{{\frac{1}{{\sqrt {n\left( {n + 1} \right)} }}}}{{\frac{1}{n}}} = \lim\limits_{n \to \infty } \frac{n}{{\sqrt {n\left( {n + 1} \right)} }} = \lim\limits_{n \to \infty } \frac{{\sqrt {{n^2}} }}{{\sqrt {n\left( {n + 1} \right)} }} = \lim\limits_{n \to \infty } \sqrt {\frac{{{n^2}}}{{{n^2} + n}}} = \lim\limits_{n \to \infty } \sqrt {\frac{{\frac{{\cancel{n^2}}}{{\cancel{n^2}}}}}{{\frac{{\cancel{n^2}}}{{\cancel{n^2}}} + \frac{\cancel{n}}{{{n^{\cancel{2}}}}}}}} = \lim\limits_{n \to \infty } \sqrt {\frac{1}{{1 + \frac{1}{n}}}} = 1.\]

As the series \(\sum\limits_{n = 1}^\infty {\left| {{a_n}} \right|} \) \(= \sum\limits_{n = 1}^\infty {\frac{1}{{\sqrt {n\left( {n + 1} \right)} }}}\) is divergent, then the initial alternating series is conditionally convergent.

Page 1 Page 2