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\begin{align} &\sum_{i=0}^n \frac{n!}{n^n} \\ \end{align}

I tried Ratio test which gives $\lim_{n\to \infty}(\frac{n}{n+1})^n$. But I'm not sure how to show this is less than 1.

mfaczz
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  • Geetha has given you the tools for the ratio test. Alternatively you can use Stirling's formula to bound the summands. – Severin Schraven May 26 '25 at 06:39
  • Thanks for the help. Would I be correct in saying that this is absolutely convergent as well since the abs(n!/n) is just still just n!/n and this series we have just proved to be convergent by the Ratio Test? @geetha290krm – mfaczz May 26 '25 at 06:55
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    There is no difference between absolute convergence and convergence for series of positive terms. – Kavi Rama Murthy May 26 '25 at 07:14
  • When $n$ th term has a factorials or factors, u can Use ratio test. – Guruprasad May 28 '25 at 03:37

1 Answers1

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The limit you are considering converges to $1/e<1$. Indeed: $$\lim_{n \to \infty} \left(\frac{n}{n+1}\right)^n = \lim_{n \to \infty} \frac{1}{\left(1+\frac{1}{n}\right)^n}=\frac{1}{e}$$

theBmax
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