Answer
Convergent
Work Step by Step
Let $f(x)=\frac{1}{x^2+6x+13}$.
This function is continuous and decreasing on $[1,\infty)$.
Evaluate $\int_1^\infty f(x) dx$:
$\int_1^\infty\frac{1}{x^2+6x+13}dx=\int_1^\infty \frac{1}{(x+3)^2+2^2}dx$
$=\frac{1}{2}\tan^{-1}(\frac{x+3}{2})]_1^\infty$
$=\frac{1}{2}\cdot \frac{\pi}{2}-\frac{1}{2}\cdot \frac{\pi}{4}$
$=\frac{\pi}{8}$
So, the integral $\int_1^\infty f(x)dx$ is convergent.
Using The Integral Test for Series, $\sum_{n=1}^\infty \frac{1}{n^2+6n+13}$ is convergent.
