How do you find the sum of Sigma 1/(k^2+1) where k is [0,3]?

Aug 12, 2017

${\sum}_{k = 0}^{3} \frac{1}{{k}^{2} + 1} = \frac{9}{5}$

Explanation:

We seek:

${\sum}_{k = 0}^{3} \frac{1}{{k}^{2} + 1} = \frac{1}{{0}^{2} + 1} + \frac{1}{{1}^{2} + 1} + \frac{1}{{2}^{2} + 1} + \frac{1}{{3}^{2} + 1}$
$\text{ } = \frac{1}{0 + 1} + \frac{1}{1 + 1} + \frac{1}{4 + 1} + \frac{1}{9 + 1}$
$\text{ } = \frac{1}{1} + \frac{1}{2} + \frac{1}{5} + \frac{1}{10}$
$\text{ } = \frac{9}{5}$