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What is the limit of the following : $$ \lim _{n \rightarrow \infty} \left[\frac{1 \cdot 1 !+2 \cdot 2 !+\cdots+n \cdot n !}{(n+1) !}\right]^{(n+1)!} $$ I think it is clear that the numerator approaches infinity faster than the denominator so the result should be $+\infty$?

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    $\begingroup$ Recall that $1 \cdot 1! + 2 \cdot 2! + ... + n \cdot n! = (n+1)!-1$. $\endgroup$ Jan 24 '21 at 5:08
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[Note] : Notice that $n\cdot n! = (n+1)!-n!$, $\forall n\in\mathbb{N}$ go ahead and try to prove that using induction. It follows that : \begin{align*} \lim _{n \to \infty} \left[\displaystyle\frac{1 \cdot 1 !+2 \cdot 2 !+\ldots+n \cdot n !}{(n+1) !}\right]^{(n+1)!}&=\lim_{n\to\infty}\left[\frac{(n+1)!-1}{(n+1)!}\right]^{(n+1)!}\\ &=\lim_{n\to\infty}\left(1-\frac{1}{(n+1)!}\right)^{(n+1)!}\\ &=\boxed{\frac{1}{e}} \end{align*}

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  • $\begingroup$ Same time, same idea ! Cheers :-) $\endgroup$ Jan 24 '21 at 5:11
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    $\begingroup$ indeed for math is a wonderful language, cheers! $\endgroup$ Jan 24 '21 at 5:14
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You have $$a_n=\Bigg[\frac {\sum_{i=1}^n i\,i!}{(n+1)!}\Bigg]^{(n+1)!}=\Bigg[\frac {(n+1)!-1}{(n+1)!}\Bigg]^{(n+1)!}=\Bigg[1-\frac {1}{(n+1)!}\Bigg]^{(n+1)!} $$ $$\log(a_n)=(n+1)! \log\Bigg[1-\frac {1}{(n+1)!}\Bigg]\sim -1 $$ $$a_n \sim \frac 1e$$

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    $\begingroup$ Thank you very much for your answer as well!! $\endgroup$ Jan 24 '21 at 5:24

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