how to compute a series whose terms are a rational function times an exponential function? How can I compute the following series?
\begin{equation}
\sum_{n=1}^\infty\frac{n+10}{2n^2+5n-3}\left(-\frac{1}{3}\right)^n
\end{equation}
I manipulated the term and got
\begin{equation}
\frac{n+10}{2n^2+5n-3}\left(-\frac{1}{3}\right)^n = \left(\frac{11}{14n-7}-\frac{1}{n+3}\right)\left(-\frac{1}{3}\right)^n\text{.}
\end{equation}
But I don't know what to do next.
 A: $$\frac{n+10}{2n^2+5n-3}=\frac{3}{2 n-1}-\frac{1}{n+3}$$
Consider the first sum
$$S_1=\sum_{n=1}^\infty\frac{x^n}{2 n-1}=y\sum_{n=1}^\infty\frac{y^{2n-1}}{2 n-1}=y\tanh ^{-1}(y)=\sqrt{x} \tanh ^{-1}\left(\sqrt{x}\right)$$ With $x=-\frac 13$ we then have
$$S_1=-\frac{\pi }{6 \sqrt{3}}$$
For the second sum
$$S_2=\sum_{n=1}^\infty\frac{x^n}{ n+3}=\frac 1 {x^3}\sum_{n=1}^\infty\frac{x^{n+3}}{ n+3}=-\frac 1 {x^3}\left(\log(1-x)+x+\frac {x^2}2 +\frac {x^3}3\right)$$
With $x=-\frac 13$ we then have
$$S_2=\frac{47}{6}-27\log \left(\frac{4}{3}\right)$$
For the total
$$\frac{47}{6}-\frac{\pi }{2 \sqrt{3}}-27\log \left(\frac{4}{3}\right)$$
A: $$S=\sum_{n=1}^\infty\frac{n+10}{2n^2+5n-3}\left(-\frac{1}{3}\right)^n$$
$$S=\sum_{n=1}^{\infty} \left(\frac{3}{2n-1}-\frac{1}{n+3}\right)(-3^{-1})^{n}=S_1-S_2$$
$$S_2=-27\sum_{n=1}^{\infty} \frac{(-3^{-1})^{n+3}}{n+3}$$
Use $\ln(1+z)=\sum_{k=1}^{\infty} (-1)^{k-1} \frac{z^k}{k}$
$$=-27\sum_{m=4}^{\infty}\frac{(-1)^m(3^{-1})^m}{m}=\frac{-47}{6}+27\ln(4/3)$$
Next use $\tan^{-1} x=\sum_{k=1}^{\infty} (-1)^{k-1}\frac{z^{2k-1}}{2k-1}$
$$S_1=\sqrt{3}\sum_{n=1}^{\infty} (-1)^n \frac{(1/\sqrt{3})^{2n-1}}{2n-1}=-\sqrt{3} \tan^{-1}(1/\sqrt{3})=\frac{-\pi}{2 \sqrt{3}}$$
Finally $$S=\frac{-\pi}{2 \sqrt{3}}+\frac{47}{6}-27\ln(4/3)$$
