A problem of infinite series

Assume that it is an infinite sum.

$\displaystyle\sum_{k=1}^{\infty} \frac{1}{(k+2)^2+k}= \displaystyle\sum_{k=1}^{\infty} \frac{1}{k^2+5k+4}= \displaystyle\sum_{k=1}^{\infty} \frac{1}{(k+1)(k+4)}= \frac{1}{3} \displaystyle\sum_{k=1}^{\infty} \frac{3}{(k+1)(k+4)}$

$=\frac{1}{3} \displaystyle\sum_{k=1}^{\infty} \frac{(k+4)-(k+1)}{(k+1)(k+4)} = \frac{1}{3} ( \sum_{k=1}^{\infty} \frac{1}{(k+1)} - \sum_{k=1}^{\infty} \frac{1}{(k+4)} )$

$= \frac{1}{3} \cdot (\frac{1}{2}+\frac{1}{3}+\frac{1}{4} )= \frac{13}{36}$

The sum is $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{(n+2)^2 + n}$ , or $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^2 + 5n + 4}$ . Note that $n^2 + 5n + 4 = (n+1)(n+4)$ , we could use the method of partial fractions in order to find that $\dfrac{1}{n^2+5n+4} = \dfrac{1}{3} \left(\dfrac{1}{n+1} - \dfrac{1}{n+4}\right)$ , so

$\begin{aligned} \sum_{n=1}^\infty \dfrac{1}{n^2+5n+4} &= \dfrac13 \sum_{n=1}^\infty \left(\dfrac{1}{n+1} - \dfrac{1}{n+4}\right) \\ &= \dfrac13\left(\dfrac12 - \dfrac15 + \dfrac13 - \dfrac16 + \dfrac14 - \dfrac17 + \dfrac15 - \dfrac18 +\dfrac16 - \dfrac19 + \ldots\right) \end{aligned}$

Some of the terms cancel, leaving only $\dfrac13\left(\dfrac12 + \dfrac13 + \dfrac14\right) = \dfrac{13}{36}$