$\text{Find the value of}$ $$\lim_{n\to \infty}\left(\sum_{r=1}^{n}\left(\frac{2r+5}{r^2+r}\right)\left(\frac{3}{5}\right)^{r+1}\right)$$
$\text{Answer}: \frac{9}{5}$
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.
Best Answer
Your idea was right:$$\sum_{r=1}^{\infty}\left(\frac{5}{r}-\frac{3}{r+1}\right)\left(\frac{3}{5}\right)^{r+1}$$ $$=\sum_{r=1}^{\infty}\left(\frac{5}{r}\right)\left(\frac{3}{5}\right)^{r+1}-\sum_{r=1}^{\infty}\left(\frac{3}{r+1}\right)\left(\frac{3}{5}\right)^{r+1}$$ $$=3\sum_{r=1}^{\infty}\left(\frac{1}{r}\right)\left(\frac{3}{5}\right)^{r}-3\sum_{r=2}^{\infty}\left(\frac{1}{r}\right)\left(\frac{3}{5}\right)^{r}$$ $$ =3\cdot \underbrace{\frac{3}{5}}_{r=1}=\frac{9}{5} $$