Prove that $\lim_{a \to \infty} \sum_{n=1}^{\infty} \frac{(n!)^a}{n^{an}} = 1$.

Question

The above sum (without the $\lim$ notation) is convergent $\forall a \in \Bbb{N}^+$, because:
$$
\sum_{n=1}^{\infty} \frac{(n!)^a}{n^{an}} = \sum_{n=1}^{\infty} \frac{n! n! \dots n!}{n^n n^n \dots n^n} \stackrel{\quad \text{because} \\ \forall n \in \Bbb{N}^+ \ n! \le n^n}{\le} \sum_{n=1}^{\infty} \frac{n!}{n^n} \approx \\ \stackrel{\quad \text{Stirling-} \\ \text{approximation}}{\approx}
\sum_{n=1}^{\infty} \frac{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}{n^n} = \sum_{n=1}^{\infty} \frac{\sqrt{2\pi n}}{e^n} = \sqrt{2\pi} \sum_{n=1}^{\infty}\frac{\sqrt{n}}{e^n} = \\ = \sqrt{2\pi} \ Li_{-\frac{1}{2}}\left(\frac{1}{e}\right) \approx 1.7728
$$
It is also decreasing $\forall a \in \Bbb{N}^+$:
$$\sum_{n=1}^{\infty} \frac{(n!)^a}{n^{an}} = \sum_{n=1}^{\infty} \left(\frac{(n!)^{a-1}}{n^{(a-1)n}}\right)\frac{n!}{n^n} \le \sum_{n=1}^{\infty} \frac{(n!)^{a-1}}{n^{(a-1)n}}$$
Because the left side is multiplied by a term that is always $\le 1$, namely $$\forall n \in \Bbb{N}^+ \quad \frac{n!}{n^n} \le 1$$
So the limit exists:
$$\exists \lim_{a \to \infty} \sum_{n=1}^{\infty} \frac{(n!)^a}{n^{an}} = ? \in \Bbb{R}^+_0$$
I examined the sum in Python and wrote a code to calculate it up to $a=20$, each time adding up the sum up to $n=1000$:

import math

def sum_term(n,a):
    return (pow(math.factorial(n),a))/pow(n,a*n)

for a in range(1,21):
    value = 0
    for n in range(1,1001):
        value += sum_term(n,a)
    print("a = " + str(a) + " | lim = " + str(value))

And its output was:

a = 1 | lim = 1.879853862175259
a = 2 | lim = 1.3099287490030924
a = 3 | lim = 1.1368584537249211
a = 4 | lim = 1.065018132743388
a = 5 | lim = 1.0317992491522754
a = 6 | lim = 1.0157461094449747
a = 7 | lim = 1.0078393253936435
a = 8 | lim = 1.0039122029986458
a = 9 | lim = 1.0019544471210995
a = 10 | lim = 1.0009768562327848
a = 11 | lim = 1.000488346517213
a = 12 | lim = 1.0002441551281933
a = 13 | lim = 1.0001220735353726
a = 14 | lim = 1.0000610358724382
a = 15 | lim = 1.0000305177372775
a = 16 | lim = 1.0000152588244295
a = 17 | lim = 1.0000076294023905
a = 18 | lim = 1.0000038146990122
a = 19 | lim = 1.000001907349021
a = 20 | lim = 1.0000009536744026

It is pretty convincing that the limit tends to $1$, however, I want to prove this mathematically. Is there a way to do this with perhaps the Squeeze theorem or other methods?

Answer 1

Hint: Note that $n!/n^n\le 1/n$ for all $n.$ Thus your sum equals $1+ R(a),$ where

$$0<R(a)\le \sum_{n=2}^{\infty}\left (\frac{1}{n}\right)^a.$$

Thus all you need to show is that $R(a)\to 0$ as $a\to \infty.$