Answer
$R_{10}\leq\frac{\pi}{2}-arctan10$
$0.982\leq S\leq1.081$
Work Step by Step
$\Sigma_{n=1}^{\infty}\frac{1}{n^{2}+1}$, using $n=10$
Find the partial sum $S_{10}$ by plugging it into a calculator:
$S_{10}=\Sigma_{n=1}^{10}\frac{1}{n^{2}+1}\approx0.9817928223$
$S_{10}=\frac{1}{2}+\frac{1}{5}+\frac{1}{10}+\frac{1}{17}+\frac{1}{26}+\frac{1}{37}+\frac{1}{50}+\frac{1}{65}+\frac{1}{82}+\frac{1}{101}\approx0.9817928223$
The function $\frac{1}{n^{2}+1}$ is positive, continuous, and ultimately decreasing. Therefore we can use the integral test to find the approximation of maximum error for the function.
$R_{10}\leq\int_{10}^{\infty}\frac{1}{n^{2}+1}dn$
$R_{10}\leq\lim\limits_{a \to \infty}\int_{10}^{a}\frac{1}{n^{2}+1}dn$
$u=n$
$du=dn$
$a=1$
$R_{10}\leq\lim\limits_{a \to \infty}[arctann]_{10}^{a}$
$R_{10}\leq\lim\limits_{a \to \infty}[arctana-arctan10]=arctan\infty-arctan10=\frac{\pi}{2}-arctan10$
$S=0.9817928223+\frac{\pi}{2}-arctan10\approx1.081461475$
