# Convergence of series with trigonometric function and factorial

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?

• $n!/n^n$ does tend to $0$, and hence your serie converges absolutely, and hence converges – Thinking Jan 28 '19 at 19:03
• By the way, partial sums of $\sin n^2$ are not bounded. – RRL Jan 28 '19 at 19:07

Hint $$\frac{n!}{n^n}=\frac{1}{n}\cdot\frac{2}{n}\cdot...\cdot\frac{n}{n}\leq \frac{1}{n}\cdot\frac{2}{n}\cdot 1 \cdot 1 \cdot... \cdot 1=\frac{2}{n^2}$$
• So if $\frac{n!}{n^n} \leq \frac{2}{n^2}$ that means that $\frac{n!}{n^n} \leq \frac{1}{n}$ and $\frac{n!}{n^n} \rightarrow 0$. So I can use Dirichlet's test? Is that right? – VakiPitsi Jan 28 '19 at 19:19
• No nevermind Dirichlet's test cause as RRL mention the partial sum is not bounded But we can say that since $\left | sin(x^2) \right | \leq1\Rightarrow \left |\frac{n!}{n^n}sin(x^2) \right |\leq \frac{1}{n^2}$ and thus the series converges right? – VakiPitsi Jan 28 '19 at 19:25
• I wasn't the downvoter, but did you mean $\frac 2 {n} . 1$ rather than $\frac 1 {n} . 1$ ? – J. W. Tanner Jan 28 '19 at 20:00
Hint: if $$k$$ is the integer part of $$n/2$$, then $$\frac{n!}{n^n} \leq \frac{k!}{n^k} \leq (k/n)^k \leq 2^{-k}.$$