Any suggestion/hint, not the whole solution, how to determine convergence/divergence of $$ \sum_{n=1}^{\infty}\dfrac{e^n \cdot n!}{n^n} $$ I'm currently stuck.
Asked
Active
Viewed 1,407 times
5
-
3You can take log. – Shine Aug 03 '14 at 14:55
-
1"convergence/divergence" of what? Do you want to find a limit or a sum? – m0nhawk Aug 03 '14 at 15:02
-
Is this supposed to be a sum? What have you tried thus far? – DrkVenom Aug 03 '14 at 15:18
-
1It is at this stage not clear whether you are asking about the sequence or the series. You can get relevant information in either case by googling Stirling Formula. One can also handle the question from basic principles. – André Nicolas Aug 03 '14 at 15:20
-
From tags it seems that the question is about series... – Ruslan Aug 03 '14 at 15:31
-
2I don't get why people think this could ever be about series. Perhaps in english one doesn't talk about convergence of sequences, I don't know. In my native language the only possible interpretation for this question is if the sequence in the question has a finite limit. – Git Gud Aug 03 '14 at 15:34
-
2See this. – David Mitra Aug 03 '14 at 16:11
-
NicolasSorry, it is series – Nick Aug 03 '14 at 16:25
-
With the question as it currently stands, Stirling's Approximation is really the easiest way to go - there are elementary approaches that get the first-order term (which is all you need here), but Stirling's is well-enough known that I wouldn't have qualms about using it for a problem where it's so clearly intended. David's link (and the item linked in turn) really cover everything about this Q as far as I can tell. – Steven Stadnicki Aug 03 '14 at 16:32
4 Answers
3
$$\ln \frac{n!e^n}{n^n}=\ln n! +n-n\ln n$$ and $$\ln n!=\sum_{k=2}^n\ln k>\sum_{k=2}^n\int_{k-1}^k\ln x\mathrm{d}x=\int_1^n\ln x\mathrm{d}x=x\ln x-x|_1^n=n\ln-n+1$$ so that $$\ln \frac{n!e^n}{n^n}>1$$ and the series diverges, since its $n$th term does not go to $0$ as $n\to\infty.$
saulspatz
- 53,131
0
It looks like the sequence is decreasing: try $\displaystyle\frac{a_n}{a_{n+1}}$.
Quang Hoang
- 15,854
-
But to show $a_n<a_{n+1}$, you only need that to be $<1$. Also, my calculation returns a little different answer. – Quang Hoang Aug 03 '14 at 15:11
-
0
OK, apologies, I take my words back about Stirling's approximation. In T. Worsch's article 'Lower and upper bounds on the binomial coefficients' it is shown that $n!$ can be lower-bounded by $(\epsilon>0)$ $$ n! > \frac{1}{1+\epsilon} \bigg(\frac{n}{e}\bigg)^n\sqrt{ 2 \pi n} $$ If you use this lower bound on the sum you have got, you'll immediately get the divergence of the series.
Alex
- 19,262