Chapter 11 - Section 11.3 - The Integral Test and Estimates of Sums - 11.3 Exercises - Page 726: 17

Convergent

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