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
 A: The reason that the approach in the OP is flawed is that the Fourier series for $\sin(x)$ for $x\in [0,\pi/2]$ is discontinuous at the end points.  This is Gibb's Phenomenon.  
In fact, we have
$$\sin(x)= \frac2\pi +\frac4\pi\sum_{n=1}^\infty \frac{\cos(4nx)}{1-16n^2}+\frac{16}{\pi}\sum_{n=1}^\infty\frac{n\sin(4nx)}{1-16n^2}$$
for $x\in (0,\pi/2)$ pointwise, but the convergence is $L^2[0,\pi/2]$.  We do not have pointwise convergence at the end points.  This is not surprising given $\sin(0)=0\ne \sin(\pi/2)=1$.
And since $\sin(0)=0$ and $\sin(\pi/2)=1$ we have
$$\frac{1+0}{2}=\frac2\pi +\frac4\pi\sum_{n=1}^\infty \frac{1}{1-16n^2} \tag 1$$
Solving $(1)$ the series of interest yields
$$\sum_{n=0}^\infty \frac{1}{1-16n^2} =1+\frac\pi4\left(\frac{1}{2}-\frac{2}{\pi}\right)=\frac{4+\pi}{8}$$
as expected!
A: 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 $\color{red}{\frac{1}{2}+\frac{\pi}{8}}$.
A: Maybe it is interesting to see that this series can be easily calculated with the residue theorem. Observing that $$S=\sum_{n\geq0}\frac{1}{1-16n^{2}}=\frac{1}{2}\sum_{n\in\mathbb{Z}}\frac{1}{1-16n^{2}}+\frac{1}{2}$$ and using $$\sum_{n\in\mathbb{Z}}f\left(n\right)=-\sum\left\{ \textrm{Res}\left(\pi\cot\left(\pi z\right)f\left(z\right)\right)\textrm{ at }f\left(z\right)\textrm{'s poles}\right\} $$ and observing that we have poles at $z=\pm\frac{1}{4}$ we get $$S=\color{red}{\frac{\pi}{8}+\frac{1}{2}}$$ as wanted.
