Strange sum

$T_{n} = \frac{1}{n^{2} + 4n}$

$= \frac{1}{n(n + 4)}$

$= \frac{1}{4}(\frac{1}{n} - \frac{1}{n + 4})$

thus we see here the series goes till infinity , and the terms $\frac{1}{5} , \frac{1}{6} ....$ will only be omitted

thus the sum is equal to

$\frac{1}{4}( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4})$

$= \frac{25}{48}$

If we denote the sum by $S$ we may observe that

$S = \sum_{n=1}^{\infty} \frac{1}{n^{2}+4n} = \sum_{n=1}^{\infty} \frac{1}{n(n+4)} = \frac{H_{4}}{4}$

where $H_{4}$ is the 4th Harmonic Number, where the k-th harmonic number may be defined as

$H_{k} = \sum_{n=1}^{k} \frac{1}{n}$

Therefore we may conclude that

$S = \frac{p}{q} = \frac{\sum_{n=1}^{4} \frac{1}{n}}{4} = \frac{25}{48}$

$\therefore p + q = 73$

More generally, though,

$S(k) = \sum_{n=1}^{\infty} \frac{1}{n(n+k)} = \frac{H_{k}}{k}$