Are these true:
$$1.\sum_{n=1}^{\infty}\frac{1}{n^2+88n}=\frac{1}{88}\sum_{n=1}^{88}\frac{1}{n}$$
$$2.\sum_{n=1}^{\infty}\frac{1}{n^2+4n}=\frac{25}{48}$$
For first one, it's factored. ¿It's not factored correctly? Because the bound is from $\infty$ to $88$ and if you multiply $\frac{1}{8}$ with the other fraction, it's not the same.
For second one, it's basic finding if it's converge. I find pattern $\frac{1}{5}+\frac{1}{12}+\frac{1}{21}$, etc.
$$\frac{1}{n^2 + 88n} = \frac{1}{n(n+88)} = \frac{1}{88}\left( \frac{1}{n} - \frac{1}{n+88} \right).$$ Therefore, $$\sum_{n=1}^\infty \frac{1}{n^2 + 88n} = \frac{1}{88} \sum_{n=1}^\infty \frac{1}{n} - \frac{1}{n+88} = \frac{1}{88} \sum_{n=1}^{88} \frac{1}{n}.$$
The other sum is handled in a similar fashion.