#### Answer

Convergent

#### Work Step by Step

$\Sigma \frac{1}{n^2+4}$
$\int_{1}^{\infty}\frac{1}{x^2+4}dx$
$u=\frac{x}{2}$
$dx=2du$
$\int_{1}^{\infty}\frac{1}{x^2+4}dx=\frac{1}{2}\int_{\frac{1}{2}}^{\infty}\frac{1}{u^2+1}du=[\frac{1}{2}tan^{-1}(u)]_{\frac{1}{2}}^{\infty}$
$\lim\limits_{t \to \infty}[\frac{1}{2}tan^{-1}(u)]_{\frac{1}{2}}^{t}=\frac{1}{2}(\frac{\pi}{2}-tan^{-1}(\frac{1}{2}))$
Convergent