3
$\begingroup$

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

or

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?

$\endgroup$
1
  • $\begingroup$ $$\left\{\left(1+\frac{1}{n}\right)^n\right\}_{n\geq 1}$$ 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
5
$\begingroup$

Hint: Show that

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

for all positive integers $n$.

$\endgroup$
7
  • $\begingroup$ 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
2
$\begingroup$

HINT

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 $\log(1+x)<x$

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

$\endgroup$
15
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.