Solve $\sum_{k=0}^{\infty}\frac{1}{1-16k^2}$

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!

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 $\color{red}{\frac{1}{2}+\frac{\pi}{8}}$.

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.