Using either root test or ratio test. I have the feeling that it is the root test, I'm not sure how to proceed from this:
$$ \sqrt[n]{n! \over n^n}= {(n!)^{1\over n} \over n} $$
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2If the terms involve factorial, always try the ratio test first. Root test is good for those cases where the terms are like $n^n, n^{n^2}$ etc or their combination. – anumosh Apr 17 '15 at 01:49
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This question, again. Will find duplicate – Thomas Andrews Apr 17 '15 at 02:03
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1possible duplicate of Finding the limit of $\frac {n}{\sqrt[n]{n!}}$ – Thomas Andrews Apr 17 '15 at 02:04
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We have the following identity: $\quad\displaystyle\sum_{n=0}^\infty\frac{n!}{n^n} ~=~ \int_0^\infty\frac{E(x)}{e^x}~dx,\quad$ where $\quad E(x)~=~\displaystyle\sum_{n=0}^\infty\frac{x^n}{n^n}\quad$ and $e^x~=~\displaystyle\sum_{n=0}^\infty \frac{x^n}{n!}~,\quad$ and $\quad\displaystyle\lim_{n\to0}n^n~=~1$. – Lucian Apr 17 '15 at 06:55
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possible duplicate of Is $\sum\limits_{n=1}^{\infty}\frac{n!}{n^n}$ convergent? – metamorphy Mar 29 '23 at 04:07
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You can use the ratio test. Here is a start
$$ \frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}}\frac{n^n}{n!} = \frac{1}{(1+1/n)^n}\longrightarrow_{n\to \infty} \frac{1}{e} <1 $$
science
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so $a_{n+1}\text{~}\dfrac{a_n}{e}$ - can we use $1+1/e+1/e^2+1/e^3+\dots$ somewhere – JMP Apr 17 '15 at 02:20
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Also $\frac{n\cdot(n-1)\cdots 2\cdot 1}{n\cdot n \cdots n} \le \frac{2\cdot 1}{n\cdot n} = \frac{2}{n^2}$ for large $n,$ if that helps
zhw.
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