# Testing the convergence of $\sum\limits_{n=1}^ \infty \frac{n!\, \pi^n}{e^{n^2}}$

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

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?

• By definition, $(n+1)!/n! = n+1$, which appears in the third line from the bottom. – Wizact Jan 7 '20 at 22:52
• @Wizact I see - thanks! Is there something like factorial for addition? Like summing terms 1-n, besides for summation notation? – Burt Jan 7 '20 at 22:54
• 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) – Wizact Jan 7 '20 at 22:57
• 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. – Gribouillis Jan 7 '20 at 23:06

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.