Answer
$n\geq1000$
Work Step by Step
$\Sigma_{n=1}^{\infty}\frac{1}{n^{2}+1}$
The function $\frac{1}{n^{2}+1}$ is positive, continuous, and ultimately decreasing. Therefore we can use the integral test to find the minimum value of $R_{n}\leq0.001$:
$R_{n}\leq\int_{n}^{\infty}\frac{1}{n^{2}+1}dn$
$R_{n}\leq\lim\limits_{a \to \infty}\int_{n}^{a}\frac{1}{n^{2}+1}dn$
$u=n$
$du=dn$
$a=1$
$R_{n}\leq\lim\limits_{a \to \infty}[arctann]_{n}^{a}$
$R_{n}\leq\lim\limits_{a \to \infty}[arctana-arctann]=arctan\infty-arctann=\frac{\pi}{2}-arctann$
$\frac{\pi}{2}-arctann\leq0.001$
$-arctann\leq\frac{1}{1000}-\frac{\pi}{2}$
$arctann\geq-\frac{1}{1000}+\frac{\pi}{2}$
$tan(arctann)\geq tan(-\frac{1}{1000}+\frac{\pi}{2})$
Round up to the nearest integer:
$n\geq1000$
