$\sum_{k=1}^\infty \frac{k!\sin(k)}{k^k}.$ convergence/divergence

Solution 1:

Remark that : $$\begin{align} \left|\dfrac{k! \sin k}{k^k}\right| &\leq \dfrac{k!}{k^k} \\&= \dfrac{1 \times 2 \times \cdots \times k}{k \times k \times \cdots \times k} \\&= \dfrac{1 \times 2}{k \times k} \, \dfrac{3 \times 4 \times \cdots \times k}{k \times k \times \cdots \times k} \\&\leq \dfrac{2}{k^2}\end{align} $$ and the series : $$\sum \dfrac{2}{k^2}$$ converges.

Solution 2:

The ratio test gives the value $1/e$ for $\lim a_{n+1}/a_n$.

So the series $$\sum_{n=1}^\infty \frac{n! \cdot 1}{n^n}$$ is convergent. Seems you just miscalculated.

So the given series

$$\sum_{k=1}^\infty \frac{k! \cdot \sin(k)}{k^k}.$$ is absolutely convergent.