<s>[INST] Finding $\sum_{r=1}^{\infty}\left(\frac{2r+5}{r^2+r}\right)\left(\frac{3}{5}\right)^{r+1}$ 
$\text{Find the value of}$ $$\lim_{n\to \infty}\left(\sum_{r=1}^{n}\left(\frac{2r+5}{r^2+r}\right)\left(\frac{3}{5}\right)^{r+1}\right)$$
$\text{Answer}: \frac{9}{5}$

Firstly I split "linear-upon-quadratic" term:
$$\frac{2r+5}{r^2+r}=\frac{5}{r}-\frac{3}{r+1}
\\ =\frac{2}{r}+3\left(\frac {1}{r} -\frac {1}{r+1}\right)$$
If it hadn't been for the $\left(\frac 35\right)^{r+1}$ term, the above step would have been very useful - splitting the single summation into two and summing individually.
Unfortunately, that's not the case. I'm unable to proceed further, though my gut says telescoping is the way.
Thanks in advance.
 [/INST] Your idea was right:$$\sum_{r=1}^{\infty}\left(\frac{5}{r}-\frac{3}{r+1}\right)\left(\frac{3}{5}\right)^{r+1}$$
$$=\sum_{r=1}^{\infty}\left(\frac{5}{r}\right)\left(\frac{3}{5}\right)^{r+1}-\sum_{r=1}^{\infty}\left(\frac{3}{r+1}\right)\left(\frac{3}{5}\right)^{r+1}$$
$$=3\sum_{r=1}^{\infty}\left(\frac{1}{r}\right)\left(\frac{3}{5}\right)^{r}-3\sum_{r=2}^{\infty}\left(\frac{1}{r}\right)\left(\frac{3}{5}\right)^{r}$$
$$
=3\cdot \underbrace{\frac{3}{5}}_{r=1}=\frac{9}{5}
$$
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