Does this series converges? Ratio Test, Factorial Numerator and exponential Denominator

$\displaystyle \sum_{k=0}^\infty \frac{A k!}{B k^k}$ where $A, \, B > 0$
Does this series converges? If A is different than B, then it obviously converges with the ratio test. But with $ A = B $. How would I know it converges?
So being more precise, does $\displaystyle \sum_{k=0}^\infty \frac{k!}{k^k}$ converges? ( Because in this way, the ratio test gives $1^-$, which is inconclusive )

Solution 1:

If $A, B > 0$ then the sum converges. Namely, it's a serie defintely non-negative. Let: $$a_n=\frac{A}{B}\cdot\frac{n!}{n^n}$$ We can use the root citeria. We have: $$\lim_{n\to +\infty}\sqrt[n]{\frac{A}{B}\cdot\frac{n!}{n^n}}=\lim_{n\to +\infty}\sqrt[n]{\frac{A}{B}}\cdot\lim_{n\to +\infty}\sqrt[n]{\frac{n!}{n^n}}$$

In my previous post, I've showned that:

$$\lim_{n\to +\infty}\sqrt[n]{\frac{n!}{n^n}}=\frac{1}{e}$$

So:

$$\lim_{n\to +\infty}\sqrt[n]{\frac{A}{B}\cdot\frac{n!}{n^n}}=\lim_{n\to +\infty}\sqrt[n]{\frac{A}{B}}\cdot\lim_{n\to +\infty}\sqrt[n]{\frac{n!}{n^n}}=0^+$$

Thus, the series converges.

Solution 2:

Yes, the sum converges. It can be proved using the direct comparison test.

We can prove that for $k>5$, $$\frac1{k^2} > \frac{k!}{k^k}$$ and since the sum over $1/k^2$ converges, the same can be said for $k!/k^k$.

Hope that's convincing.