Sum inside integral

Since $\frac{x^2}{x^2 + 1} = 1 - \frac{1}{x^2 + 1}$ , we now integrate this expression first to obtain:

$\int_{1}^{n} 1 - \frac{1}{x^2 + 1} dx = x - arctan(x)|_{1}^{n} = n - arctan(n) - 1 + \frac{\pi}{4}.$

Now focusing on the sum:

$\Sigma_{n=0}^{\infty} (n + \frac{\pi}{4} - 1) - arctan(n)$

one notices by the Comparison Test that;

$n < n + \frac{\pi}{4} - 1 \Rightarrow \boxed{\Sigma_{0}^{\infty} n < \Sigma_{0}^{\infty} n + \frac{\pi}{4} - 1}$ (i)

Since the LHS of (i) is divergent, the RHS is as well.