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.
 A: Note that
$$\sum_{n=1}^\infty \frac {r^n}n=\sum_{n=1}^\infty \int_0^x r^{n-1}dr=\int_0^x\sum_{n=1}^\infty r^n dr=\int_0^x \frac 1{1-r}dr=\bigg[-\ln(1-r)\bigg]_0^x=\ln\left(\frac 1{1-x}\right)$$
and
$$\sum_{n=1}^\infty\frac {r^n}{n+2}=\sum_{n=3}^\infty \frac {r^{n-2}}{n}=\frac 1{r^2}\sum_{n=3}^\infty\frac {r^n}n=\frac 1{r^2}\left[\left(\sum_{n=1}^\infty\frac {r^n}n\right)-r-\frac {r^2}2\right]=\frac 1{r^2}\left[\ln\left(\frac 1{1+r}\right)\right]-\frac 1r-\frac 12$$
Putting $r=\dfrac 34$ gives
$$\sum_{n=1}^\infty \frac{\left(\frac 34\right)^n}n=\ln 4$$
and $$\sum_{n=1}^\infty \frac {\left(\frac 34\right)^n}{n+2}=\frac {16}9\ln 4-\frac {11}6$$
Hence 
$$\begin{align}
\sum_{n=1}^\infty \frac {7n+32}{n(n+2)}\cdot \left(\frac 34\right)^n
&=16\sum_{n=1}^\infty \frac{\left(\frac 34\right)^n}{n} 
-9\sum_{n=1}^\infty \frac{\left(\frac 34\right)^n}{n+2} \\
&=16\;\;\ln4\;\;\;\;-\;\;9\left(\frac {16}9 \ln4 -\frac {11}6\right)\\
&=\color{red}{\frac{33}2}
\end{align}$$
A: Hint: for $|x|<1$ the power series $f(x):=\sum_{n=1}^{\infty}\frac{1}{n}x^n$ is convergent.
Hence $f'(x)=\sum_{n=1}^{\infty}x^{n-1}=\frac{1}{1-x}$ for $|x|<1$.
A: 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:
$$\sum_{n=1}^{\infty} (-1)^{n-1} \, \left(\frac{7 \, n +32}{n(n+2)}\right) \, \left( \frac{3}{4}\right)^{n} = \frac{15}{2}.$$
A: HINT:
$$7n+32=4^2(n+2)-3^2\cdot n$$
$$\implies\dfrac{7n+32}{n(n+2)}\left(\dfrac34\right)^n=\dfrac{4^2(n+2)-3^2\cdot n}{n(n+2)}\left(\dfrac34\right)^n=\dfrac{16\left(\dfrac34\right)^n}n-\dfrac{16\left(\dfrac34\right)^{n+2}}{(n+2)}$$
