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This is the explanation to the answer to the question: Test the convergence of the series $\displaystyle\sum\limits_{n=1}^ \infty \frac{n!\, \pi^n}{e^{n^2}}$

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I don't understand what they did with the factorial terms. How did they disappear? It seems to me that $(n+1)!$ is much larger than $n!$ - which will change the limit to approach infinity. Can someone help?

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    $\begingroup$ By definition, $(n+1)!/n! = n+1$, which appears in the third line from the bottom. $\endgroup$ – Wizact Jan 7 '20 at 22:52
  • $\begingroup$ @Wizact I see - thanks! Is there something like factorial for addition? Like summing terms 1-n, besides for summation notation? $\endgroup$ – Burt Jan 7 '20 at 22:54
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    $\begingroup$ Yes, you can show (e.g. by induction) that 1+2+3+...+n = n(n+1)/2. (check it for a couple of small values) $\endgroup$ – Wizact Jan 7 '20 at 22:57
  • $\begingroup$ Note that Stirling's fornula also implies that $(n! e^{-n^2})^{1/n}\longrightarrow 0$, which proves the convergence thanks to the Cauchy-Hadamard theorem. $\endgroup$ – Gribouillis Jan 7 '20 at 23:06
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I don't understand what they did with the factorial terms. How did they disappear?

This is simply because, by definition, $$n!=1\cdot2\cdot3\dotsm(n-1)\cdot n $$ and therefore, $$\frac{(n+1)!}{n!}=\frac{\color{red}{1\cdot2\cdot3\dotsm(n-1)\cdot n}\cdot(n+1)}{\color{red}{1\cdot2\cdot3\dotsm(n-1)\cdot n}}=n+1$$ since the red terms cancel out exactly.

It seems to me that $(n+1)!$ is much larger than $n!$ - which will change the limit to approach infinity.

In some sense your intuition is right, in absolute terms $|(n+1)!-n!|$ is very large and does grow to infinity rapidly. However, here we have the ratio $(n+1)!/n!=n+1$, which still grows to infinity, but in a relatively tame way---it is only linear. The limit as $n\to\infty$ happens to be $0$ here because although we have a factor that grows linearly to infinity, the $e^{2n+1}$ factor in the denominator is much more significant (as it is an exponential term), and hence dominates the linear term and forces the expression overall to tend to $0$. This is perhaps a nice illustration of why not to trust your initial instincts before you've worked out the algebra and can confidently see what's going on.

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