Context:
I was randomly putting infinite sums of this form ($k\in\mathbb N)$ in online calculators. $$\sum_{n\ge1}\left(\frac{n!}{n^n}\right)^k$$
I noticed that as $k$ increases, the sums becomes more and more closer to $1$.
Question: Can we prove if $$\lim_{k\to\infty}\sum_{n\ge1}\left(\frac{n!}{n^n}\right)^k = 1 $$
Thanks !
How about the following: Let $f_k$ be the sequence $((\frac{n!}{n^n})^k)_{n=1}^\infty$ and let $\mu$ be the counting measure on $\mathbb{N}$. Then $$ \int_\mathbb{N} f_k d\mu = \sum_{k=1}^\infty f_k(n), $$ see Integration with respect to counting measure.
Now $0 \le f_k \le f_1 \in L^1(\mu)$ and $f_k(1) =1 \to 1=:g(1)$, $f_k(n) \to 0=:g(n)$ $(n\ge 2)$ for $k \to \infty$. By Lebesgue's Dominated Convergence Theorem $$ \lim_{k \to \infty}\int_\mathbb{N} f_k d\mu = \int_\mathbb{N} g d\mu=1. $$
Edit: To avoid this abstract setting you can do the following. Fix $\varepsilon>0$ and let $n_0>1$ be such that $$ \sum_{n=n_0+1}^\infty \frac{n!}{n^n} < \varepsilon/2. $$ Choose $k_0$ such that $$ |1-\sum_{n=1}^{n_0}(\frac{n!}{n^n})^{k_0}|= \sum_{n=2}^{n_0}(\frac{n!}{n^n})^{k_0} < \varepsilon/2. $$ Note that $f_k(n)$ is decreasing in $k$. Hece for $k \ge k_0$ $$ |1-\sum_{n=1}^{\infty}(\frac{n!}{n^n})^{k}| \le |1-\sum_{n=1}^{n_0}(\frac{n!}{n^n})^{k}| + \sum_{n=n_0+1}^\infty \frac{n!}{n^n} < \varepsilon. $$
A more elementary approach would be the following:
Notice the the series starts with
$$\sum_{n\ge 1}\left(\frac{n!}{n^n}\right)^k = 1^k + \left(\frac12\right)^k + \left(\frac29\right)^k + \ldots$$
The terms $\frac{n!}{n^n}$ are shrinking fast, so the goal is to show that everything after $1^k$ will sum to smaller and smaller values when $k$ increases.
We have for the quotient of two consecutive terms:
$$\frac{(n+1)!}{(n+1)^{n+1}}/\frac{n!}{n^n} = \frac{(n+1)!n^n}{n!(n+1)^{n+1}} = \frac{(n+1)n^n}{(n+1)^{n+1}} = \frac{n^n}{(n+1)^n} = \left(\frac{n}{n+1}\right)^n = \left(1-\frac1{n+1}\right)^n.$$
The last term is is decreasing when $n$ increases (and will unltimately converge to $e^{-1}$), but what we need is simply that it is at most $\frac12$ for $n \ge 1$.
That means we get
$$\frac{n!}{n^n} \le \left(\frac12\right)^{n-1},$$
as it's true with equality for $n=1,2$, then for higher $n$ the quotient between one term and the next from the left hand side will be smaller than $\frac12$.
So we get
$$\sum_{n\ge 1}\left(\frac{n!}{n^n}\right)^k \le 1^k + \sum_{n\ge 2}\left(\left(\frac12\right)^{n-1}\right)^k = 1 + \sum_{n\ge 1}\left(\left(\frac12\right)^n\right)^k = 1 + \sum_{n\ge 1}\left(\left(\frac12\right)^k\right)^n.$$
The sum at the end here is a simple geometrical series, with starting value and quotient equal to $\left(\frac12\right)^k$. This can be calculated exactly, and we finally get
$$\sum_{n\ge 1}\left(\frac{n!}{n^n}\right)^k \le 1 + \sum_{n\ge 1}\left(\left(\frac12\right)^k\right)^n = 1 + \frac{\left(\frac12\right)^k}{1-\left(\frac12\right)^k}.$$
The limit if that quotient for $k\to\infty$ is easy to find, the enumarator tends to $0$, while the denominator tends to $1$, so the whole qutient tends to $0$.
That means $\lim_{k\to\infty}\sum_{n\ge 1}\left(\frac{n!}{n^n}\right)^k \le 1.$ That was the hard part. As can be seen from the very first formula, the sum $\sum_{n\ge 1}\left(\frac{n!}{n^n}\right)^k$ starts with first summand $1$, and only positive summands follow, so we obviously have $\lim_{k\to\infty}\sum_{n\ge 1}\left(\frac{n!}{n^n}\right)^k \ge 1.$
That concludes the proof.
$$F_k=\sum_{n=1}^\infty\left(\frac{n!}{n^n}\right)^k=1+\sum_{n=2}^\infty\left(\frac{n!}{n^n}\right)^k$$
For a short cut, using Stirling approximation $$\frac{n!}{n^n} \sim \sqrt{2 \pi n} \,e^{-n}$$ Consider $$G_k=1+(2\pi)^{\frac k 2}\sum_{n=2}^\infty n^{\frac k 2}\,e^{-n k}$$
$$\sum_{n=2}^\infty n^{\frac k 2}\,e^{-n k}=\text{Li}_{-\frac{k}{2}}\left(e^{-k}\right)-e^{-k}$$
Asymptotically $$\log \left(\text{Li}_{-\frac{k}{2}}\left(e^{-k}\right)-e^{-k} \right) \sim -\frac 5 3 k$$
$$G_k \sim 1+\exp\left(-\frac{10-3 \log (2 \pi )}{6} k \right)\sim 1+ e^{-3 k/4}$$
