Compute $\sum_{n=1}^\infty (\frac34)^n \frac{7n+32}{n(n+2)}$

Question

Question: Compute $$\sum_{n=1}^\infty \left(\frac34\right)^n \frac{7n+32}{n(n+2)}$$

I first did the partial fraction decomposition into: $$\sum_{n=1}^\infty \left(\frac34\right)^n \frac{16}n - \sum_{n=1}^\infty \left(\frac34\right)^n \frac9{n+2}$$ Here I am stuck because I wrote out a couple terms and I don't see a way that terms can cancel out. Can I get some help?

Answer 1

There is no need to cancel terms across both series because of the $(3/4)^n$. Evaluating both series individually (using the Maclaurin series of $\log(1-x)$) gives $$S=\frac{16}{\log4}-\left(-\frac{33}2+\frac{16}{\log4}\right)=\frac{33}2$$

Answer 2

In this this type of problems just do telescoping now make the second terms in your partial fraction in powers of n+2 that is just divide and multiply by 16 in second term then do telescoping and possible answers are 33/2

Answer 3

Hint:

Let $\dfrac{7n+32}{n(n+2)}\cdot\left(\dfrac34\right)^n=f(n)-f(n-2)$

where $f(m)=\dfrac a{m+2}\cdot\left(\dfrac34\right)^m$

so that for $r\ge4,$ $$\sum_{n=2}^r\dfrac{7n+32}{n(n+2)}\cdot\left(\dfrac34\right)^n=\sum_{n=2}^r(f(n)-f(n-2))=f(r)+f(r-1)-f(1)-f(0)$$

$$f(n)-f(n-2)=\left(\dfrac34\right)^n\left(\dfrac a{n+2}-\dfrac an\cdot\dfrac{16}9 \right)$$

We need $\dfrac{7n+32}{n(n+2)}=\dfrac a{n+2}-\dfrac an\cdot\dfrac{16}9=-a\cdot\dfrac{7n+32}{9n(n+2)}\implies a=-9$