Calc6.16

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Discuss the convergence of \sum _{{n=2}}^{{\infty }}{\frac  {(-1)^{n}}{\ln n}} including whether the sum converges absolutely or conditionally.

This series converges by the alternating series test. The terms alternate and go to 0 and the absolute value of the terms decreases.

By definition a series converges absolutely if the series with terms that are the absolute values of the terms of the original series also converges. In this case, the series we need to consider is

\sum _{{n=2}}^{{\infty }}\left|{\frac  {(-1)^{n}}{\ln n}}\right|=\sum _{{n=2}}^{{\infty }}{\frac  {1}{\ln n}}\,

One way to determine the convergence of this new series is the direct comparison test.

{\frac  {1}{\ln n}}>{\frac  {1}{n}}\,

We know that the series \sum _{{n=2}}^{{\infty }}{\frac  {1}{n}} diverges. Thus, by the direct comparison test, since the terms of our series are all even greater than a series which diverges, our series must also diverge.

We say \sum _{{n=2}}^{{\infty }}{\frac  {(-1)^{n}}{\ln n}} converges conditionally because the series itself converges but the series \sum _{{n=2}}^{{\infty }}\left|{\frac  {(-1)^{n}}{\ln n}}\right| diverges.


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