Convergence of series with trigonometric function and factorial

Question

I'm stuck on this.. Given $\sum_{n=1}^{\infty}\frac{n!}{n^n}\cdot sin(n^2)$ I have to determine if it's convergent or not.

I can see that $sin(n^2)$ is bounded and so are the partial sums of it. But $\frac{n!}{n^n}$ doesnt converge to $0$ to use Dirichlet's test

Also my intuition is that it's not convergence but I can't find another divergent sequence to use the comparison test

Any hints?

Answer 1

Hint $$\frac{n!}{n^n}=\frac{1}{n}\cdot\frac{2}{n}\cdot...\cdot\frac{n}{n}\leq \frac{1}{n}\cdot\frac{2}{n}\cdot 1 \cdot 1 \cdot... \cdot 1=\frac{2}{n^2}$$

Answer 2

Hint: if $k$ is the integer part of $n/2$, then $$\frac{n!}{n^n} \leq \frac{k!}{n^k} \leq (k/n)^k \leq 2^{-k}.$$