Calc6.17

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Discuss the convergence of \sum _{{n=1}}^{{\infty }}{\frac  {(-1)^{{n+1}}}{n^{2}+3n+4}} including whether the sum converges absolutely or conditionally.

We can use the alternating series test here to determine that the series converges since the terms go to 0 and the absolute value of the terms decreases. However, this series converges absolutely and any series that converges absolutely simply converges so let us just prove that this series is absolutely convergent. To do this, examine the series

\sum _{{n=1}}^{{\infty }}\left|{\frac  {(-1)^{{n+1}}}{n^{2}+3n+4}}\right|=\sum _{{n=1}}^{{\infty }}{\frac  {1}{n^{2}+3n+4}}\,

Comparing this series to another, we see that

{\frac  {1}{n^{2}+3n+4}}<{\frac  {1}{n^{2}}}\,

and we know that the series \sum _{{n=1}}^{{\infty }}{\frac  {1}{n^{2}}} converges. Thus, by the direct comparison test, the series \sum _{{n=1}}^{{\infty }}\left|{\frac  {(-1)^{{n+1}}}{n^{2}+3n+4}}\right|=\sum _{{n=1}}^{{\infty }}{\frac  {1}{n^{2}+3n+4}} converges and thus our original series \sum _{{n=1}}^{{\infty }}{\frac  {(-1)^{{n+1}}}{n^{2}+3n+4}} not only converges but converges absolutely.


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