(Solved): n=1(1)n1arctan(n)n=1nesin(n) Determine absolute canvergeace, or condi ...

$\begin{array}{l}{\sum }_{n=1}^{\infty }(-1{)}^{n-1}\arctan (n)\\ {\sum }_{n=1}^{\infty }\frac{\sin (n)}{n^e}\end{array}$ Determine absolute canvergeace, or conditimelly convergent, or divergent ${\sum }_{n=1}^{\infty }(-1{)}^{n-1}\left(\frac{n^2}{n^2+4}\right)$ Determine convergent or divergent $\begin{array}{l}{\sum }_{n=1}^{\infty }(-1{)}^{n-1}\arctan (n)\\ {\sum }_{a=1}^{\infty }\frac{\sin (n)}{n^e}\end{array}$ Determine absolute caavergence, or conditim. Ily ceavergent, of divergeat ${\sum }_{n=1}^{\infty }(-1{)}^{n-1}\left(\frac{n^2}{n^2+4}\right)$ Use Ratio test to determine convegnt or divege- ${\sum }_{n=0}^{\infty }\frac{n^{100}100^n}{(n+100)!}$