<s>[INST] Sum of $\frac{9}{5^2\cdot 2 \cdot 1}+\frac{13}{5^3 \cdot 3 \cdot 2}+\frac{17}{5^4 \cdot 4 \cdot 3}+\ldots \text{to} \hspace{0.1 cm} \infty $ Find the sum of the series 
$$\frac{9}{5^2\cdot 2 \cdot 1}+\frac{13}{5^3 \cdot 3 \cdot 2}+\frac{17}{5^4 \cdot 4 \cdot 3}+\ldots \text{to}  \hspace{0.1 cm} \infty $$
I see that $r_{th}$ term is $\frac{5+4r}{5^{r+1} \cdot r \cdot(r+1)}$ and I am look to find sum of $n$ terms and then put $n \to \infty$ . 
Could someone provide me insight to find some of $n$ terms or any other better approach?
 [/INST] $$\frac{5+4r}{5^{r+1} r(r+1)} = \frac{5(1+r)-r}{5^{r+1}r(r+1)} = \frac{1}{ 5^{r} r}-\frac{1}{5^{r+1}(r+1)}$$
hence the series is telescopic:
$$ \sum_{r\geq 1}\frac{5+4r}{5^{r+1} r(r+1)} = \frac{1}{5^1\cdot 1}=\color{red}{\frac{1}{5}}$$
and:
$$ \sum_{r=1}^{R}\frac{5+4r}{5^{r+1} r(r+1)} = \frac{1}{5}-\frac{1}{5^{R+1}(R+1)}.$$
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