<s>[INST] Telescoping(?) an infinite series Find the value of the sum $\displaystyle \sum\limits_{n=1}^{\infty} \frac{(7n+32) \cdot3^n}{n(n+2) \cdot 4^n}.$
Using partial fraction decomposition, I found the above expression is equivalent to $\displaystyle \sum\limits_{n=1}^{\infty} \frac{25}{n} \cdot \left(\frac{3}{4}\right)^n - \sum\limits_{n=1}^{\infty} \frac{18}{n+2} \cdot \left(\frac{3}{4}\right)^n,$ where I got couldn't find a closed form of either expression because of the $\left(\dfrac{3}{4}\right)^n.$
Similarly, trying to telescope one of the the sums with $\displaystyle \sum\limits_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+2}\right) \cdot \left(\frac{3}{4}\right)^n$ fails for the same reason. How can I further simplify the above expression? Thanks.
 [/INST] Consider:
$$\frac{7 \, n +32}{n(n+2)} = \frac{7(n+2) + 9}{n(n+2)} = \frac{16}{n} - \frac{9}{n+2}$$
from which
\begin{align}
\sum_{n=1}^{\infty} \frac{7 \, n +32}{n(n+2)} \, t^{n} &= - 16 \, \ln(1-t) - \frac{9}{t^2} \, \left( \sum_{n=1}^{\infty} \frac{t^{n}}{n} - t - \frac{t^{2}}{2} \right) \\
&= \left(\frac{9}{t^{2}} - 16 \right) \, \ln(1-t) + \frac{9}{t} + \frac{9}{2}.
\end{align}
When $t = 3/4$ then this becomes
\begin{align}
\sum_{n=1}^{\infty} \frac{7 \, n +32}{n(n+2)} \, \left( \frac{3}{4}\right)^{n} = \frac{33}{2}. 
\end{align}
It becomes evident that the only two values of $t$ for which the logarithmic term has a zero coefficient are $t = \pm (3/4)$. In this view it is developed:
$$\sum_{n=1}^{\infty} (-1)^{n-1} \, \left(\frac{7 \, n +32}{n(n+2)}\right) \, \left( \frac{3}{4}\right)^{n} = \frac{15}{2}.$$
</s>