Does the series $\sum\limits_{n=1}^\infty\frac{\sin(n)n!}{n^n}$ converge?

Question

$\sum\limits_{n=1}^\infty\frac{\sin(n)n!}{n^n}$

Please let me know how you did it. Thank you.

Answer 1

Note that $xy$ with $x+y$ fixed is maximized when $x=y$. Thus $$n!=(1\cdot n)(2\cdot(n-1))\cdots(\lfloor (n+1)/2\rfloor\cdot\lceil (n+1)/2\rceil)<\left(\frac{n+1}{2}\right)^n=\frac{(n+1)^n}{2^n}$$ and since $$\left(\frac{n+1}{n}\right)^n=\left(1+\frac{1}{n}\right)^n<e$$ we have $$\sum\limits_{n=1}^\infty\left|\frac{\sin(n)n!}{n^n}\right|<\sum_{n=1}^\infty \frac{|\sin(n)|(n+1)^n}{2^nn^n}<\sum_{n=1}^\infty\frac{1}{2^n}\cdot e=e$$ thus the sum converges absolutely, so converges.