Prove the absolute convergence of the series?

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1 Answer
Bill K.
Dec 30, 2017

The Ratio Test is the best thing to use here.

Explanation:

To prove that #sum_{n=1}^{infty}((-100)^{n})/(n!)# converges absolutely, we must show that #sum_{n=1}^{infty}|((-100)^{n})/(n!)|=sum_{n=1}^{infty}(100^{n})/(n!)# converges.

This is best done by the Ratio Test. Let #a_{n}=100^{n}/(n!)#. Then #a_{n+1}=(100*100^{n})/((n+1)*n!)# and #a_{n+1}/a_{n}=100/(n+1)->0=L# as #n->infty#.

Since #L<1#, it follows that #sum_{n=1}^{infty}a_{n}=sum_{n=1}^{infty}(100^{n})/(n!)# converges. Hence, #sum_{n=1}^{infty}((-100)^{n})/(n!)# converges absolutely.

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