Telescoping(?) an infinite series

Question

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.

Answer 1

HINT:

$$7n+32=4^2(n+2)-3^2\cdot n$$

$$\implies\dfrac{7n+32}{n(n+2)}\left(\dfrac34\right)^n=\dfrac{4^2(n+2)-3^2\cdot n}{n(n+2)}\left(\dfrac34\right)^n=\dfrac{16\left(\dfrac34\right)^n}n-\dfrac{16\left(\dfrac34\right)^{n+2}}{(n+2)}$$

Answer 2

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 $|x|<1$.

Answer 3

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: $$\sum_{n=1}^{\infty} (-1)^{n-1} \, \left(\frac{7 \, n +32}{n(n+2)}\right) \, \left( \frac{3}{4}\right)^{n} = \frac{15}{2}.$$

Answer 4

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

$$\begin{align} \sum_{n=1}^\infty \frac {7n+32}{n(n+2)}\cdot \left(\frac 34\right)^n &=16\sum_{n=1}^\infty \frac{\left(\frac 34\right)^n}{n} -9\sum_{n=1}^\infty \frac{\left(\frac 34\right)^n}{n+2} \\ &=16\;\;\ln4\;\;\;\;-\;\;9\left(\frac {16}9 \ln4 -\frac {11}6\right)\\ &=\color{red}{\frac{33}2} \end{align}$$