Calc6.18

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Discuss the convergence of \sum _{{n=1}}^{{\infty }}{\frac  {(-1)^{n}}{n!}} including whether the sum converges absolutely or conditionally.

The terms of the factorial function get very big very quickly. So we would think this sequence would converge. Let’s look at the series which contains the absolute value of the terms of this sequence.

\sum _{{n=1}}^{{\infty }}\left|{\frac  {(-1)^{n}}{n!}}\right|=\sum _{{n=1}}^{{\infty }}{\frac  {1}{n!}}

For n>3, we have

{\frac  {1}{n^{2}}}>{\frac  {1}{n!}}

and we know that \sum _{{n=1}}^{{\infty }}{\frac  {1}{n^{2}}} converges because it is a p-series with p>1. Thus, the series \sum _{{n=1}}^{{\infty }}{\frac  {1}{n!}} also converges by the direct comparison test. The first three terms do not change the convergence of a series. Adding a finite number to a convergent series does not change it to a divergent series. Finally, since this converges, we know that \sum _{{n=1}}^{{\infty }}{\frac  {(-1)^{n}}{n!}} converges absolutely.


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