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

Solution 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 $a\to \infty.$

Solution 2:

Perhaps here is the approach. First term of your sum is 1, and all others are in $(0,1)$. So if we exchange the two limits, $x_n^a \to 0$ as $a \to \infty$, so the entire sum will be zero.

You have to check the axioms if the limits can be exchanged though, but everything looks safe -- both positive and convergent...

Solution 3:

It is elementary that $$\frac{n!}{n^n}\le\frac12$$ for all $n>1$ and the series is squeezed in

$$\left[1,\frac1{1-\dfrac1{2^a}}\right].$$

For large $a$, is is approximately

$$\dfrac1{1-\dfrac1{2^a}}$$ as witnessed by

$$\dfrac1{1-\dfrac1{2^{20}}}=\color{green}{1.00000095367}522590\cdots$$