Calc6.15

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Discuss the convergence of the series \sum _{{n=1}}^{{\infty }}{\frac  {(-1)^{{n+1}}n^{2}}{e^{n}}}\,.

The absolute value of the terms of this sequence is decreasing and goes to 0 as n\rightarrow \infty . If you can not see this, use L'H{\hat  {o}}pital's Rule. First, in taking the limit as n\rightarrow \infty , we get {\frac  {\infty }{\infty }}. Since this is an indeterminate form, it is likely that L'H{\hat  {o}}pital's Rule might be the way to find the limit of the terms.

\lim _{{n\to \infty }}{\frac  {n^{2}}{e^{n}}}=\lim _{{n\to \infty }}{\frac  {2n}{e^{n}}}=\lim _{{n\to \infty }}{\frac  {2}{e^{n}}}=0\,

Thus, by L'H{\hat  {o}}pital's Rule, since the limit exists, this method gives us the correct limit. The terms of this series go to 0 because the exponential function increases much more quickly than the power function. This also indicates that the absolute value of the terms is decreasing.

Therefore, by the alternate series test, we know that this series converges.


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