3

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?

J.G.
  • 115,835
Ghost
  • 1,105
  • $$\left{\left(1+\frac{1}{n}\right)^n\right}_{n\geq 1}$$ is an increasing sequence converging to $e$, hence all the terms of your series are greater than one. – Jack D'Aurizio Oct 27 '18 at 04:23

2 Answers2

5

Hint: Show that

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

for all positive integers $n$.

Carl Schildkraut
  • 32,931
  • 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 $e$? – Ghost Oct 26 '18 at 20:39
  • @Ghost That sounds like a good idea. – Carl Schildkraut Oct 26 '18 at 20:40
  • Either $\lim_{n\rightarrow \infty}(1+\tfrac{1}{n})^n$ or $\sum_{k=0}^{\infty}\frac{1}{k!}$ would be your definition of $e$. – OgvRubin Oct 26 '18 at 20:42
  • @CarlSchildkraut I wish I could have approved 2 answers since both of them helped me solve the problem. Thanks for all the help! – Ghost Oct 26 '18 at 21:46
  • @Ghost In that case I think that Carl deserves that more than me! – user Oct 26 '18 at 21:48
  • @gimusi If you are ok with that I will give the green mark to him. For me, both of you helped me alot, so thanks for that! – Ghost Oct 26 '18 at 21:50
  • @Ghost Yes of course! I fully agree with that :) – user Oct 26 '18 at 21:51
2

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 $\log(1+x)<x$

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

user
  • 154,566
  • Ahm, sorry if I ask but did you use Taylor expansion? – Ghost Oct 26 '18 at 20:45
  • @Ghost Yes but we can also use the simpler $\log(1+x)<x$. I add that. – user Oct 26 '18 at 20:46
  • 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. – Ghost Oct 26 '18 at 20:47
  • @Ghost Once we know $\left(1+\frac{1}{n}\right)^n < e$ we can evaluate the limit. – user Oct 26 '18 at 20:50
  • 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. – TonyK Oct 26 '18 at 20:50
  • @TonyK What can we deduce from $$\left(1+\frac{1}{n}\right)^n < e\implies \left(1+\frac{1}{n}\right)^{n^2} < ?$$ and what is is $x$ in $$\frac{e^n}{(1+\frac{1}{n})^{n^2}}<x$$ – user Oct 26 '18 at 20:52
  • In the end we get that $\frac{e^n}{(1+\frac{1}{n})^{n^2}}>1$ am I right? – Ghost Oct 26 '18 at 21:02
  • @Ghost Exactly! That's the argument suggested by Carl Schildkraut! My first one was a overkilling! What about the limit then? – user Oct 26 '18 at 21:04
  • @gimusi I guess that if the apply the limit to the inequality we get that $\lim_{n->\infty}\frac{e^n}{(1+\frac{1}{n})^{n^2}}\geq1$ so we found out that it can not be $0$ so the series is divergent. – Ghost Oct 26 '18 at 21:06
  • @Ghost Very nice! You got it and very good your claim that $\lim_{n->\infty}\frac{e^n}{(1+\frac{1}{n})^{n^2}}\geq1$ and not $>1$. – user Oct 26 '18 at 21:08
  • @Ghost As a more interestin question now, what about $$\sum \frac{(1+\frac{1}{n})^{n^2}}{e^n}$$ – user Oct 26 '18 at 21:09
  • Thanks. Still, it would be nice if I would find out the exact value of that limit also. – Ghost Oct 26 '18 at 21:09
  • @Ghost In that case Taylor's expansion can be helpful! – user Oct 26 '18 at 21:11
  • @gimusi I will try to also try the series of the inverse expression and send you a private message with what I have done and what are the results. Thanks for all the help! – Ghost Oct 26 '18 at 21:45
  • @Ghost You are welcome! I'll take a look to that. – user Oct 26 '18 at 21:52