Chapter 11 - Section 11.3 - The Integral Test and Estimates of Sums - 11.3 Exercises - Page 725: 7

Divergent

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