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I have this succession's limit: $ \lim_{n \to +\infty}{\tfrac{n^{n!}}{(n!)^n}} $ . I've been tried by using root's criterion and I've got $ +\infty $. Can someone show me how to solve this exercise? Just to compare it with mine. Thanks

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  • $\begingroup$ Do you know Stirling approximation ? $\endgroup$ – anonymus Feb 18 '17 at 11:25
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Probably not the answer expected in the excercise but you can do just rough estimates and it will work in this case. For example from $n!^n\leq (n^n)^n=n^{n^2}$ we have $$\frac{n^{n!}}{n!^n}\geq \frac{n^{n!}}{n^{n^2}}=n^{n!-n^2}\geq n \to \infty$$ since $n!$ grows much faster than $n^2$ (shouldn't be hard to show $n!-n^2 \geq 1$ for $n \geq 4$).

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Hint.

You have

$$\frac{n^{n!}}{(n!)^n}=e^{n!\log(n)-n\log(n!)}=e^{n((n-1)!\log(n)-\log(n!))}.$$

And recall the fact that

$$\log(n!)=\sum_{k=1}^n \log(k).$$

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The numerator has $n!$ factors all equaling $n$, while the denominator has just $n^2$ of them, most smaller than $n$.

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