Decide convergence of this series

Question

How to prove the series $$\sum_{n=1}^\infty \frac {e^n\cdot n!}{n^n}$$ diverges? I tried D'Alambert and result is 1 and I'm stuck with Raabe.

The second answer here may help. – David Mitra Jun 26 '14 at 11:20 — David Mitra, Jun 26 '14 at 11:20

score 4 · Answer 1 · answered Jun 26 '14 at 11:08

4

Hint: use $$ n! \approx \sqrt{2\pi n} \left( \frac{n}{e} \right)^n $$ and deduce that the $n$-th term...

answered Jun 26 '14 at 11:08

Siminore

35,136

I can't use approximations, just standard tests like here:http://www.math.hawaii.edu/~ralph/Classes/242/SeriesConvTests.pdf – Meow Jun 26 '14 at 11:11
2

I'm too old to pretend that Stirling's formula does not exist :-( – Siminore Jun 26 '14 at 11:14
I understand =D thank you anyway – Meow Jun 26 '14 at 11:18
This looks like an overkill for such a simple problem... – Alex Jun 26 '14 at 12:33
I don't always like "elementary" approaches. – Siminore Jun 26 '14 at 12:41

score 2 · Answer 2 · answered Jun 26 '14 at 11:18

2

Can you use this little trick? $ \exp(n) = 1 + n + n^2/2 +\cdots + n^n/n! + \cdots $. We then obtain that $\exp(n) n! = n! + n n! \cdots + n^n + \cdots > n^n$.

answered Jun 26 '14 at 11:18

Leo

1,215

score 1 · Answer 3 · answered Jun 26 '14 at 11:19

1

Hint

Let us define $$a_n=\frac {e^n\cdot n!}{n^n}$$ So, $$\frac{a_{n+1}}{a_n}=e \left(\frac{n}{n+1}\right)^n$$ which is number greater than $1$.

I am sure that you can take from here.

answered Jun 26 '14 at 11:19

Claude Leibovici

260,315

@DavidMitra. I know (I hope) but I did not want to go to the limits in order the OP continues from this. Cheers :) – Claude Leibovici Jun 26 '14 at 11:22
but this equals 1, not greater :P – Meow Jun 26 '14 at 11:24
@DavidMitra. I always forget to tell you that the quotation from Leibnitz in your profile is one of my very favored. When much much younger, I used to say that I share something with Leibnitz (the naswer is $Leib$); unfortunately, that's all. Cheers :) – Claude Leibovici Jun 26 '14 at 11:25
@Mare. The ratio I gave you is strictly larger than $1$ which is its limit for an infinite value of $n$. – Claude Leibovici Jun 26 '14 at 11:27
1

@ClaudeLeibovici: I'm not sure how to use it since limit is 1 so it's inconclusive – Meow Jun 26 '14 at 11:43
Sorry for being dense before... @Mare, this shows the sequence $(a_n)$ is increasing; since $a_1>0$, the sum can't converge. See the answers here to see why $b_n=\Bigl({n\over n+1}\Bigr)^n ={1\over \bigl(1+1/n\bigr)^n }> 1/e$; and hence why $a_{n+1}/a_n >1$. Very nice Claude! – David Mitra Jun 26 '14 at 13:54

score 1 · Answer 4 · answered Jun 26 '14 at 12:23

1

One more way: log the summand to get $$ \log e^n n! - n \log n = n + \sum_{k=1}^{n} \log k - n \log n \geq 1 $$ Hence the original function is greater than $e^1>0$, hence it diverges. This is because $\sum_{k=1}^{n} \log k >\int_{1}^{n} \log x = n \log n -n +1$

answered Jun 26 '14 at 12:23

Alex

19,262

please see the edit – Alex Jun 26 '14 at 12:28
Great! thank you. This is acctualy a proof limit is infinity, which is even better! =) – Meow Jun 26 '14 at 12:30
please see the edit - the limit of the sequence itself is not infinity, it is $e$, but it means that the sum dies diverge. My mistake was not to consider the second term, which is also important. Sorry! – Alex Jun 26 '14 at 12:32
You can also check it by using Euler-Maclaurin approximation of sum with integrals up to the second term – Alex Jun 26 '14 at 16:46

Decide convergence of this series

4 Answers4