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Compute $$\lim_{n\to\infty} \left[\ln\left(\frac{1}{0!}+\frac{1}{1!}+\cdots+\frac{1}{n!}\right)\right]^n$$ If you have some nice proofs and you're willing to share them, then I thank you and you definitely have my upvote!

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There is a routine proof, the missing tail of the sum is $\lt \frac{1}{nn!}$. – André Nicolas Feb 10 '13 at 19:29
more proofs... More than what? – Did Feb 10 '13 at 19:30
@did: I'll be glad even if MSE users offer me one proof. I'm usually interested in many proofs for each problem. The beauty of math is to watch/tackle a problem from many points of view. – user 1618033 Feb 10 '13 at 19:34
@MaisamHedyelloo: yeah, right. Thanks! We get the case $1^{\infty}$. – user 1618033 Feb 10 '13 at 19:37
Why one would want to gather proofs of THIS statement is a bit mysterious to me. The limit is 1, quite crude estimates yield it, and the result is not especially interesting nor meaningful nor esthetically pleasing. But maybe I am missing something? – Did Feb 10 '13 at 19:40
up vote 8 down vote accepted

One could check that $$ \log\left(\sum\limits_{k=0}^n\frac{1}{k!}\right)=1+\alpha_n $$ where $$ \alpha_n=\log\left(1-e^{-1}\sum\limits_{k=n+1}^\infty\frac{1}{k!}\right) $$ Note that $$ 0\leq\lim\limits_{n\to\infty}n\alpha_n=\lim\limits_{n\to\infty}n(-e^{-1})\sum\limits_{k=n+1}^\infty\frac{1}{k!}\leq\lim\limits_{n\to\infty}n\frac{-1}{enn!}=0 $$ So $$ \lim\limits_{n\to\infty}\log^n\left(\sum\limits_{k=0}^n\frac{1}{k!}\right)= \lim\limits_{n\to\infty}\left(\left(1+\alpha_n\right)^{\frac{1}{\alpha_n}}\right)^{n\alpha_n}=e^0=1 $$

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Thanks for your way! (+1) – user 1618033 Feb 10 '13 at 20:11

The obvious way is to bound the missing tail of the series. That tail is less than a geometric series with sum $\frac{1}{nn!}$.

It follows that the thing $w$ inside the logarithm satisfies $$e-\frac{1}{nn!}\lt w\lt e.$$ Thus $$1+\log\left(1-\frac{1}{enn!}\right) \lt \log w \lt 1.$$ By the Taylor series for the logarithm, we have $$\log\left(1-\frac{1}{enn!}\right)=-\frac{1}{enn!}+o(1/nn!).$$ In particular, for large $n$, the logarithm is $\gt -\frac{2}{enn!}$. Now dealing with the $n$-th power is easy. The limit is $1$. There is an enormous amount of slack. A tail that is $O(1/n^2)$ would have been plenty good enough.

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thanks for both hint and answer! (+1) – user 1618033 Feb 10 '13 at 20:13

Let $$u_n = \frac{1}{0!}+\frac{1}{1!}+\cdots+\frac{1}{n!}.$$ The problem becomes to evaluating $$\ell := \lim_{n \to \infty} (\log u_n)^n$$ But since $$e - \frac {1} {n n!} < u_n < e,$$ we have $$1 - \frac {1} {e n n!} - \frac {1} {(e n n!)^2} + O \left(\frac {1} {(e n n!)^3}\right) < \log u_n < 1.$$ This gives us $$\ell = 1.$$

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