Solve $\sum_{k=0}^{\infty}\frac{1}{1-16k^2}$ I could use some help on calculating this infinite sum: $\sum_{k=0}^{\infty}\frac{1}{1-16k^2}$. Included was that I had to start with a Fourier series for the function $f:\Re \to \Re: x \mapsto \sin(x)$ for $x\in[0, \frac{\pi}{2}[$, so let's start with that.
Let 
\begin{eqnarray*}
g(x) = \frac{a_0}{2} + \sum_{k=1}^{\infty}a_k\cos(4kx) + \sum_{k=1}^{\infty}b_k\sin(4kx).
\end{eqnarray*} 
This is the Fourier series for $f$.
With $a_k = \frac{4}{\pi}\int_{0}^{\frac{\pi}{2}}\sin(x)\cos(4kx)dx$ and $b_k = \frac{4}{\pi}\int_{0}^{\frac{\pi}{2}}\sin(x)\sin(4kx)dx$. Solving this leads to (or at least I found that): $a_k = \frac{4}{(1-16k^2)\pi}$, $b_k = \frac{16k}{(1-16k^2)\pi}$ for $k\geq1$ and $a_0 = \frac{4}{\pi}$. Bringing this to $g(x)$ gives:
\begin{eqnarray*}
g(x) = \frac{2}{\pi} + \sum_{k=1}^{\infty}\frac{4}{(1-16k^2)\pi}\cos(4kx) + \sum_{k=1}^{\infty}\frac{16k}{(1-16k^2)\pi}\sin(4kx). 
\end{eqnarray*}
Since $f(x) \approx g(x)$, we can say that $f(0) = g(0)$. We get 
\begin{eqnarray*}
\sin(0) = \frac{2}{\pi} + \sum_{k=1}^{\infty}\frac{4}{(1-16k^2)\pi}\cos(0) + \sum_{k=1}^{\infty}\frac{16k}{(1-16k^2)\pi}\sin(0),
\end{eqnarray*}
this becomes 
\begin{eqnarray*}
0 = \frac{2}{\pi} + \sum_{k=1}^{\infty}\frac{4}{(1-16k^2)\pi}.
\end{eqnarray*}
We get 
\begin{eqnarray*}
\frac{-2}{\pi} = \frac{4}{\pi}\sum_{k=1}^{\infty}\frac{1}{1-16k^2}
\end{eqnarray*} so 
\begin{eqnarray*}
\sum_{k=1}^{\infty}\frac{1}{1-16k^2} = \frac{-1}{2}.
\end{eqnarray*} We need the sum from k = 0. The term $\frac{1}{1-16k^2}$ for k = 0 gives 1, so we add 1 to both sides. This leads to my solution 
\begin{eqnarray*}
\sum_{k=0}^{\infty}\frac{1}{1-16k^2} = \frac{1}{2}.
\end{eqnarray*}
However, when approaching this sum numerically and using Wolfram, I find that the sum should be $\frac{4+\pi}{8}$. Could some help and point out where I went wrong with my approach? Thanks in advance

First lets tidy the sum up a little
\begin{eqnarray*}
\sum_{k=0}^{\infty} \frac{1}{1-16k^2}=1-\sum_{k=1}^{\infty} \frac{1}{16k^2-1}
\end{eqnarray*}
Now partial fractions gives
\begin{eqnarray*}
\sum_{k=1}^{\infty} \frac{1}{16k^2-1}= \sum_{k=1}^{\infty} \left( \frac{1/2}{4k-1}-\frac{1/2}{4k+1} \right)
\end{eqnarray*}
Now use $\frac{1}{i}= \int_0^1 x^{i-1} dx $. We have
\begin{eqnarray*}
\sum_{k=1}^{\infty} \frac{1}{16k^2-1}= \frac{1}{2}\sum_{k=1}^{\infty}\int_0^1   \left( x^{4k-2}-x^{4k} \right) dx
\end{eqnarray*}
Now interchange the sum & the integral.... & perform the geometric sums
\begin{eqnarray*}
\sum_{k=1}^{\infty} \frac{1}{16k^2-1}=\frac{1}{2} \int_0^1  \left( \frac{x^2-x^4}{1-x^4} \right) dx = \frac{1}{2}\int_0^1   \left( \frac{x^2}{1-x^2} \right) dx =\frac{1}{2}(1-\frac{\pi}{4}). 
\end{eqnarray*}
Now putting it all back together ... we have $\color{red}{\frac{1}{2}+\frac{\pi}{8}}$.
