Sequences and series that I cant partial fraction decomposition I have to calculate the partial sum for an equation. How can I calculate the sum for
$$\sum_{n=1}^{\infty}\frac{1}{16n^2-8n-5}$$
And I think that is not simple as $$\sum_{n=1}^{\infty}\frac{1}{16n^2-8n-3}.$$ 
TIA
 A: Using partial fractions,
$$ \frac{1}{16 n^2 - 8n - 5} = \frac{\sqrt{6}}{48(n - (1+\sqrt{6})/4)} - \frac{\sqrt{6}}{48(n - (1-\sqrt{6})/4)}$$
and then (carefully) taking the limit of the sum of this from $1$ to $N$ as $N \to \infty$, your sum becomes
$$ \frac{\sqrt{6}}{48} \left( \Psi\left( \frac{3+\sqrt{6}}{4}\right) - \Psi\left( \frac{3-\sqrt{6}}{4}\right) \right)$$
I do not believe that this can be simplified significantly.
A: For computing partial fractions, use the one you know how to work as a model for the other. The method is the same.
The zeroes of $16n^2-8n-3$ are $\frac{8\pm\sqrt{256}}{32}$, so $\frac34$ and $-\frac14$, which means you can write the polynomial as $16(n-\frac34)(n+\frac14) = (4n-3)(4n+1)$. Your candidate for the partial fraction is then
$$\frac{A}{4n-3} + \frac{B}{4n+1}.$$
Use the same method for the other expression.
The zeroes of $16n^2-8n-5$ are $\frac{8\pm 8\sqrt{6}}{32}$, so $\frac14 + \frac14\sqrt{6}$ and $\frac14 - \frac14\sqrt{6}$, which means you can write the polynomial as
$16(n-\frac14 - \frac14\sqrt{6})(n - \frac14 + \frac14\sqrt{6}) = (4n - 1 - \sqrt{6})(4n - 1 + \sqrt{6})$. Your candidate for the partial fraction is then
$$\frac{A}{4n-1 - \sqrt{6}} + \frac{B}{4n - 1 + \sqrt{6}}.$$

Additional comment: This works for any quadratic, even if the zeroes of the polynomial are not real (!). For example,

$$x^2+1 = (x-i)(x+i)$$
so
$$\frac{1}{x^2+1} = \frac{A}{x-i}+\frac{B}{x+i}$$ would be the candidate for the partial fraction decomposition (what are $A$ and $B$ in this case?).
