Determine whether the series $$\sum_{n=0}^\infty\frac{2^{n^2}}{n!},$$ is convergent or divergent.

I know I have to use the ratio test.

  • 1
    $\begingroup$ I am afraid it is not comprehensible. $\endgroup$ – Yiorgos S. Smyrlis Nov 9 '14 at 21:21
  • $\begingroup$ is it comprehensible now? im new to this website and do not know how to format the questions $\endgroup$ – John Nov 9 '14 at 21:23
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    $\begingroup$ So, what does the ratio test tell you? $\endgroup$ – Hagen von Eitzen Nov 9 '14 at 21:25
  • $\begingroup$ It is fine. Let me fix it with LaTeX. $\endgroup$ – Yiorgos S. Smyrlis Nov 9 '14 at 21:25
  • $\begingroup$ it is actually 2 to the power of n squared in the numerator $\endgroup$ – John Nov 9 '14 at 21:29

Yes it diverges, and the simplest test to use is indeed the ratio test: $$ \frac{a_{n+1}}{a_n}=\frac{2^{(n+1)^2} n!}{2^{n^2}(n+1)!}=\frac{2^{2n+1}}{n+1}\to \infty, $$ as $n\to\infty$.

Hence the series diverges.

  • $\begingroup$ Yes, indeed, corrected now. $\endgroup$ – Yiorgos S. Smyrlis Nov 18 '14 at 23:28
  • $\begingroup$ It should be divergent. $\endgroup$ – Crostul Nov 18 '14 at 23:32
  • $\begingroup$ Corrected! Thanks! $\endgroup$ – Yiorgos S. Smyrlis Nov 18 '14 at 23:35

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