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trying to determine if the series is conditionally convergent or divergent. $$\sum_{n = 1}^\infty \frac{2^{n^{2}}}{n!}$$ with n! i tried the ratio test on the series $$\frac{2^{(n+1)^{2}}}{(n+1)!} * \frac{n!}{2^{n^{2}}} = \frac{2^{2n+1}}{(n+1)} $$ which is > 1 as $n\to \infty$ and is overall divergent ? not sure if I am on the right track.

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    $\begingroup$ You're on the right track. I believe the limit of the ratio is $\infty$. $\endgroup$ – Matthew Leingang Jun 15 '17 at 16:16
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    $\begingroup$ Side note: since all terms are positive, "conditionally convergent" is equivalent to "absolutely convergent." $\endgroup$ – Clement C. Jun 15 '17 at 16:21
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Well, applying the ratio test:

$$\lim_{\text{n}\to\infty}\space\left|\frac{\frac{2^{\left(\text{n}+1\right)^2}}{\left(\text{n}+1\right)!}}{\frac{2^{\text{n}^2}}{\text{n}!}}\right|=\lim_{\text{n}\to\infty}\space\left|\frac{2^{2\text{n}+1}}{\text{n}+1}\right|\space\to\space\infty\tag1$$

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Note that ${2^{n^2} \over n!} = { (2^n)^n \over n!} \ge { (2^n)^n \over n^n}= ({2^n \over n})^n \ge 1$. Hence $\sum_n {2^{n^2} \over n!} \ge N$ for all $N$.

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  • $\begingroup$ @ClementC.: That was embarrassing, thanks for catching that. Added another approach. $\endgroup$ – copper.hat Jun 15 '17 at 16:46

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