Q:

Is the sum of the following series a finite number or not? Explain. $ \sum_{k=1}^\infty \frac{5\sin^2k}{k!} $

Is the sum of the following series a finite number or not? Explain. 
  $$ \sum_{k=1}^\infty \frac{5\sin^2k}{k!} $$

A:

$$\begin{eqnarray*}\sum_{k=1}^{+\infty}\frac{2\sin^2 k}{k!}=\sum_{k=0}^{+\infty}\frac{1-\cos(2k)}{k!}&=&e-\text{Re}\sum_{k\geq 0}\frac{e^{2k i}}{k!}\\[0.2cm]&=&e-\text{Re}\left(\exp\left(\cos 2+i\sin 2\right)\right)\\[0.2cm]&=&\color{red}{e-e^{\cos 2}\cos(\sin 2)}.\end{eqnarray*}$$

A:

$$ 
\sum_{k=1}^{\infty} \frac{5 \sin^2(k)}{k!} \le \sum_{k=1}^{\infty} \frac{5}{k!}=5\times \sum_{k=1}^{\infty} \frac{1}{k!} =5(e-1)
$$
The series converge.

