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I'm working on a problem sheet and it ask to discuss the convergence of $$\sum \frac{n!}{{n}^{n}}$$ By D'Lembert's ratio test, $$\lim_{n->\infty}\frac{{a}_{n+1}}{{a}_{n}} = 1$$ and so, is inconclusive.

Using Cauchy's root test,

$$\lim_{n->\infty}({\frac{n!}{{n}^{n}}})^\frac{1}{n}=1$$

What are my alternatives?

Should I take the integral of the term of the series above? Would integrating factorial works?

Marc van Leeuwen
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guest
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3 Answers3

7

Actually the ratio test turns out to be conclusive :

$$\begin{align}\lim_{n\to\infty}\frac{{a}_{n+1}}{{a}_{n}} &=\lim_{n\to\infty}\dfrac{(n+1)!}{(n+1)^{n+1}} \cdot \dfrac{n^n}{n!} \\~\\&=\lim_{n\to\infty}\dfrac{n+1}{(n+1)^{n+1}} \cdot \dfrac{n^n}{1} \\~\\&=\lim_{n\to\infty} \left(\dfrac{n}{n+1}\right)^n\\~\\&=\lim_{n\to\infty} \left(\dfrac{\color{blue}{n+1}-1}{n+1}\right)^n\\~\\&=\lim_{n\to\infty} \left(\color{blue}{1}+\dfrac{-1}{n+1}\right)^n\\~\\&=e^{-1}~~\color{Red}{\star} \\~\\&\lt 1\end{align}$$

$\color{red}{\star}$ : please see $e^x$ limit definition

AgentS
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    Looks like the root test also works. – Regret Mar 19 '15 at 09:02
  • Looks like it does. I see where I went wrong although I do not see how the argument follows from the fourth line to the fifth. – guest Mar 19 '15 at 09:06
  • Ahh skipped few lines in between... I have updated, please check now – AgentS Mar 19 '15 at 09:09
  • Sorry it was a typo. I was able to follow what you've just edited. I meant to say I was not able to follow the 5th to the 6th line. – guest Mar 19 '15 at 09:11
  • Ohk.. got you :) I'll add a reference to that limit.. – AgentS Mar 19 '15 at 09:12
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    Incase if that $n+1$ in the denominator is throwing you off $$\lim_{n\to\infty} \left(\color{blue}{1}+\dfrac{-1}{n+1}\right)^n = \lim_{n\to\infty} \left(\color{blue}{1}+\dfrac{-1}{n+1}\right)^{n+1}\left(\color{blue}{1}+\dfrac{-1}{n+1}\right) \~\=\lim_{n\to\infty} \left(\color{blue}{1}+\dfrac{-1}{n+1}\right)^{n+1}~\lim_{n\to\infty} \left(\color{blue}{1}+\dfrac{-1}{n+1}\right) =e^{-1}\cdot 1 $$ – AgentS Mar 19 '15 at 09:16
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    Thanks but I was able to recognize that trick much like x = sqrt[x^(2)] "!" =P – guest Mar 19 '15 at 09:29
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$$\frac{n!}{n^n}=\frac{1}{n}\frac{2}{n}\cdots\frac{n-1}{n}\frac{n}{n}<\frac2{n^2}$$

Regret
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  • I will try this too. – guest Mar 19 '15 at 09:06
  • Ah yes it converges since if Sum[2/(n^(2)] from n=1 to infinity converges then by the direct comparison test, a_n converges – guest Mar 19 '15 at 09:14
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    How do you see that this inequality holds? – Jonas Dahlbæk Mar 19 '15 at 10:46
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    @user161825: In the middle term there, you have $\frac 1n\frac 2n$ multiplied several times by something less than $1$. Therefore, $\frac 1n\frac 2n=\frac 2{n^2}$ is greater. – Regret Mar 19 '15 at 10:50
  • I also don't think that's instantly immediate for which n this works, or why exactly $n^2$ pops up. It calls for a proof by induction. – Nikolaj-K Mar 19 '15 at 19:21
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Hint: $\dfrac{n!}{n^n} < \dfrac{2}{n^2}$.

DeepSea
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