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Test the convergence of the series $$\sum_{n=1}^\infty \frac{n^n}{3^n n!}$$

I know that if the nth term tends to $\infty$ then the series is divergent and if it is tends to 0 it is convergent . Also I'm familiar with some test e.g. Ratio test, d'Alembert test, comparison test etc. But I could not solve it in proper way. I know as $n$ increases $n^n$ increases more rapidly than $n!$ or $3^n$ but no idea when they both are multiplied

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Are you familiar with Stirling's approximation of the factorial function? – Gerry Myerson Jul 17 '12 at 9:24
If the $n$-th term tends to zero, the series does not need to converge. – Dirk Jul 17 '12 at 9:24
@Gerry yes I know the Stirling's approximation of the factorial function.Pleas explain how this can be used in this problem – Argha Jul 17 '12 at 9:45
@Arjang That's the second minor title edit to a very old post you've made in minutes. Please check the "active" date before making such small edits; more so as the original title did not suffer from lack of clarity. – Lord_Farin Jun 25 '13 at 8:26
@Lord_Farin : Thanks for the comment, the minor title edit was due to the fact the related links on the right handside was taking two lines. You can check for yourself that there are links on the right hand side that are unneccerily too long, In due time I be shortening them as well. – Arjang Jun 25 '13 at 8:56
up vote 6 down vote accepted

I don't see a problem in solving it by using D'Alembert's Ratio-Test.

You need to find the convergence of $$\sum_{n=1}^\infty \dfrac{n^n}{3^n n!}$$

So let $u_n= \dfrac{n^n}{3^n n!} $ which implies $u_{n+1}= \dfrac{(n+1)^{n+1}}{3^{n+1} (n+1)!}$.

Hence $$\dfrac{u_n}{u_{n+1}}=\dfrac{n^n}{3^n n!} . \dfrac{3^{n+1} (n+1)!}{(n+1)^{n+1}}$$ $$\dfrac{u_n}{u_{n+1}}=\dfrac{n^n}{3^n n!} . \dfrac{3^{n}.\ 3 \ .(n+1).n!}{(n+1)^{n}.(n+1)}$$

$$\dfrac{u_n}{u_{n+1}}=\dfrac{3.n^n}{(n+1)^n}$$ $$\dfrac{u_n}{u_{n+1}}=\dfrac{3}{(1+\dfrac{1}{n})^n}$$

$$\lim_{n\to \infty} \dfrac{u_n}{u_{n+1}}= \dfrac{3}{e} > 1 $$ $$( \ \rm{ Since } \ \lim_{n \to \infty } (1+\dfrac{1}{n})^n = e ) $$ See this for the proof of the term $e$.

Hence by D'Alemberts ratio test we have $l>1 ( l=\lim_{n\to \infty} \dfrac{u_n}{u_{n+1}})$ , $\sum_{n=1}^\infty \dfrac{n^n}{3^n n!}$ converges.

Thank you.

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thank you it is really helpful – Argha Jul 17 '12 at 9:50
@Ranabir : Did you really understand the step by step simplification, I missed working out many steps. So please do ask me. – Iyengar Jul 17 '12 at 9:53
Thank you @vanguard2k for your edit. – Iyengar Jul 17 '12 at 10:01
Not leaving OP very much to do. – Gerry Myerson Jul 17 '12 at 13:09

The ratio test should work, but
1. you have to do some clever manipulation to $(n+1)^{n+1}/n^n$, and
2. you have to know the value of $\lim_{n\to\infty}(1+n^{-1})^n$.

Does that help?

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I think the first term should sound like $\dfrac{ n^n}{(n+1)^{n}}$ after simplification. Thank you. Can you edit it, or can I have a chance to edit it ? , I was seeking a prior permission . – Iyengar Jul 17 '12 at 10:46
@Iyengar, I think what I wrote was what I meant to write. Why do you think it should be something else? I'm accustomed to using the ratio test in the form, $u_{n+1}/u_n$, not $u_n/u_{n+1}$. – Gerry Myerson Jul 17 '12 at 13:06
@Iyengar - I would agree with Gerry's comment about the usual way in which the ratio test is written. I was upset by your initial solution until I realized that you were using the reciprocal of the customary form. You have a good solution, but I, too, wish you had left a bit more for the questioner to do. – Chris Leary Jul 17 '12 at 14:29
@GerryMyerson : Oops, Sorry !!. I thought to "ERR IS HUMAN", and just have thought that you have made a typo. I learnt it by using $\dfrac{u_n}{u_n{n+1}}$ . Anyway with my experience I can no way give you any suggestion. My apologies. – Iyengar Jul 17 '12 at 14:38

Using D'alambert's ratio test we can get that the series $$\sum \frac{n^n}{x^n\cdot n!}$$ converges for all $x>e$ and diverges for $x<e$.

It worth mentioning that at $x=e$ the sum diverges (can be seen by using Stirling's approximation).

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