<s>[INST] How can I calculate the sum $\sum_{n=1}^{\infty} \frac{1}{16n^2-8n-3}$? I have to calculate the partial sum for an equation. How can I calculate the sum for 
$$\sum_{n=1}^{\infty} \frac{1}{16n^2-8n-3}\ \text{?}$$
Thanks.
 [/INST] You should start with partial fraction decomposition by finding the roots of the denominator\begin{align}\frac{1}{16n^2-8n-3}&=\frac{1}{\left(4n-3\right)\left(4n+1\right)}=\frac14\left(\frac{1}{4n-3}-\frac{1}{4n+1}\right)\end{align} In this way, the initial series can be written as telescoping series and therefore you can calculate the partial sum up to $N \in \mathbb N$ and then let $N\to+\infty$ as follows:
\begin{align}\sum_{n=1}^{N}&\left(\frac{1}{4n-3}-\frac{1}{4n+1}\right)=\\[0.2cm]&=\left(1-\frac15\right)+\left(\frac15-\frac19\right)+\left(\frac19-\frac1{15}\right)\dots+\left(\frac{1}{4N-3}-\frac1{4N+1}\right)\\[0.2cm]&=1+\left(\frac15-\frac15\right)+\left(\frac19-\frac19\right)+\dots+\left(\frac{1}{4N-3}-\frac1{4N-3}\right)-\frac1{N+1}\\[0.2cm]&=1-\frac{1}{4N+1}
\end{align}
So, returning to initial series
\begin{align}\sum_{n=1}^{+\infty}\frac14\left(\frac{1}{4n-3}-\frac{1}{4n+1}\right)&=\lim_{N\to+\infty}\frac14\sum_{n=1}^{N}\left(\frac{1}{4n-3}-\frac{1}{4n+1}\right)\\[0.2cm]&=\lim_{N\to+\infty}\frac{1}{4}\left(1-\frac1{4N+1}\right)=\frac14(1-0)=\frac14\end{align} 
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