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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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    @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.2013-06-25

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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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    Not leaving OP very much to do.2012-07-17
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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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    @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.2012-07-17
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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.

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