We have
[tex]\displaystyle \sum_{n=1}^\infty \frac{n}{n^2+8} < \int_1^\infty \frac{x}{x^2+8}\,\mathrm dx[/tex]
For the integral, substitute y = x ² + 8 and dy = 2x dx. Then
[tex]\displaystyle \int_1^\infty \frac{x}{x^2+8}\,\mathrm dx = \frac12 \int_9^\infty \frac{\mathrm dy}y = \frac12 \ln(y)\bigg|_{y=9}^{y\to\infty} = \infty[/tex]
The integral diverges, so the sum also diverges by the integral test.
