I need to study the convergence of the series $\sum_{n=1}^{\infty}\frac{e^n}{(1+\frac{1}{n})^{n^2}}$.

First, I was thinking of finding the limit: $\lim_{n\to\infty}\frac{e^n}{(1+\frac{1}{n})^{n^2}}$ cause if we find that it is different then $0$ the problem is over, since we know the series will be divergent. The only problem is that I do not know how to do it.

If the limit is $0$ then, I think we can do it by using the fact that if we have a series $\sum_{n=1}^{\infty}a_n$ and we can find $b_n$ so that $a_n<b_n$ then:

if $\sum_{n=1}^{\infty}b_n$ is convergent then $\sum_{n=1}^{\infty}a_n$ i convergent


if $\sum_{n=1}^{\infty}a_n$ is divergent then $\sum_{n=1}^{\infty}b_n$ is divergent

If this will not work we can try to use limit comparison test, but I doubt it will be necessary.

The main problem for me first is to find if the limit is $0$ or not.

Can you help me out to find out how to solve it?

  • $\begingroup$ $$\left\{\left(1+\frac{1}{n}\right)^n\right\}_{n\geq 1}$$ is an increasing sequence converging to $e$, hence all the terms of your series are greater than one. $\endgroup$ – Jack D'Aurizio Oct 27 '18 at 4:23

Hint: Show that

$$\left(1+\frac{1}{n}\right)^n < e$$

for all positive integers $n$.

  • $\begingroup$ I have seen this inequality soo many times still I have never tried to show it. Should I show that $(1+\frac{1}{n})^n$ is a growing sequence that converges to $e$? $\endgroup$ – Ghost Oct 26 '18 at 20:39
  • $\begingroup$ @Ghost That sounds like a good idea. $\endgroup$ – Carl Schildkraut Oct 26 '18 at 20:40
  • $\begingroup$ Either $\lim_{n\rightarrow \infty}(1+\tfrac{1}{n})^n$ or $\sum_{k=0}^{\infty}\frac{1}{k!}$ would be your definition of $e$. $\endgroup$ – OgvRubin Oct 26 '18 at 20:42
  • $\begingroup$ @CarlSchildkraut I wish I could have approved 2 answers since both of them helped me solve the problem. Thanks for all the help! $\endgroup$ – Ghost Oct 26 '18 at 21:46
  • $\begingroup$ @Ghost In that case I think that Carl deserves that more than me! $\endgroup$ – user Oct 26 '18 at 21:48


We have that

$$\left(1+\frac{1}{n}\right)^{n^2}=e^{n^2\log \left(1+\frac{1}{n}\right)}=e^{n-\frac1{2}+O\left(\frac1{n}\right)}\sim\frac{e^n}{\sqrt e}$$

or in a simple way, following the idea by Carl Schildkraut, using $\log(1+x)<x$

$$\left(1+\frac{1}{n}\right)^{n^2}=e^{n^2\log \left(1+\frac{1}{n}\right)}<e^{\left(n^2\cdot \frac1n\right)}=e^n$$

  • $\begingroup$ Ahm, sorry if I ask but did you use Taylor expansion? $\endgroup$ – Ghost Oct 26 '18 at 20:45
  • $\begingroup$ @Ghost Yes but we can also use the simpler $\log(1+x)<x$. I add that. $\endgroup$ – user Oct 26 '18 at 20:46
  • $\begingroup$ Ok, I will also try that. I do not know how to use Taylor expansion so writing down something I do not really know that it is will not help me. $\endgroup$ – Ghost Oct 26 '18 at 20:47
  • $\begingroup$ @Ghost Once we know $\left(1+\frac{1}{n}\right)^n < e$ we can evaluate the limit. $\endgroup$ – user Oct 26 '18 at 20:50
  • $\begingroup$ But $(1+1/n)^n$ tends to $e$, so we also seem to have that $(1+1/n)^{n^2}\sim e^n$. What am I missing here? I know from numerical evidence that your version is correct. $\endgroup$ – TonyK Oct 26 '18 at 20:50

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