Compute $\lim\limits_{n\to\infty} \left[\ln\left(\frac{1}{0!}+\frac{1}{1!}+\cdots+\frac{1}{n!}\right)\right]^n$

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!

Solution 1:

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 $$ \lim\limits_{n\to\infty}\log^n\left(\sum\limits_{k=0}^n\frac{1}{k!}\right)= \lim\limits_{n\to\infty}\left(\left(1+\alpha_n\right)^{\frac{1}{\alpha_n}}\right)^{n\alpha_n}=e^0=1 $$

Solution 2:

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.

Solution 3:

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 $$\ell = 1.$$