Prove $\lim_{k\to\infty}\sum_{n=0}^\infty \frac{1}{(n!)^k}=2$

Question

How to prove that $$\lim_{k\to\infty}\sum_{n=0}^\infty \frac{1}{(n!)^k}=2 \,\,?$$

A plot shows that the values seem to quickly converge to $2$.

Cannot exclude a duplicate but couldn't find it in the search.

Answer 1

Note that for $n \geq 2$ you have obviously $$\frac 1{n!} \leq \frac 1{2^{n-1}}$$

Now squeeze:

$$2 < \sum_{n=0}^{\infty}\frac 1{(n!)^k} < 2+\underbrace{\sum_{\color{\blue}{n=2}}^{\infty}\frac 1{2^{(n-1)k}}}_{=\frac 1{2^k}\sum_{n=2}^{\infty}\frac 1{2^{(n-2)k}}<\frac 1{2^k}\sum_{\color{\blue}{n=0}}^{\infty}\frac 1{2^{n}}}$$

Answer 2

Here is a demonstration, but not necessarily a rigorous proof:

$$\sum_{n=0}^\infty \frac1{n!^k}= \frac1{0!^k}+\frac1{1!^k}+\sum_{n=2}^\infty \frac1{n!^k} =2+\sum_{n=2}^\infty \frac1{n!^k}$$

Now we need to show that:

$$\lim_{k\to\infty}\sum_{n=2}^\infty \frac1{n!^k}=0 $$

For $n\ge2,n!>1$ and raising an $x>1$ to higher powers will give increasingly larger numbers, but taking the reciprocal of those increasingly larger numbers will produce numbers getting closer to $0$. Since the power here is infinite, $\lim\limits_{k\to\infty}n!^k\to\infty$ and $\frac1{n!^k}\to 0$

So $$2+\lim_{k\to\infty}\sum_{n=2}^\infty \frac1{n!^k} =2+\sum_{n=2}^\infty 0=2$$