Evaluating $\sum_{k=1}^{\infty} \frac{\cos^2(k)}{4k^2-1}$ What options do I have for this series? No idea how to do it.
$$\sum_{k=1}^{\infty} \frac{\cos^2(k)}{4k^2-1}$$
 A: Avoiding contour integration:
$$
\begin{align}
\sum_{k=1}^\infty\frac{\cos^2(kx)}{4k^2-1}
&=\sum_{k=1}^\infty\frac{\cos(2kx)+1}{2(4k^2-1)}\\
&=\frac12\mathrm{Re}\left(\sum_{k=1}^\infty\frac{e^{2ikx}+1}{4k^2-1}\right)\\
&=\frac14\mathrm{Re}\left(\sum_{k=1}^\infty\frac{e^{2ikx}+1}{2k-1}-\frac{e^{2ikx}+1}{2k+1}\right)\\
&=\frac14\mathrm{Re}\left(e^{2ix}+1+\sum_{k=1}^\infty\frac{e^{2i(k+1)x}-e^{2ikx}}{2k+1}\right)\\
&=\frac14\mathrm{Re}\left(2+(e^{2ix}-1)+(e^{2ix}-1)\sum_{k=1}^\infty\frac{e^{2ikx}}{2k+1}\right)\\
&=\frac14\mathrm{Re}\left(2+(e^{2ix}-1)\sum_{k=0}^\infty\frac{e^{2ikx}}{2k+1}\right)\\
&=\frac14\mathrm{Re}\left(2+(e^{ix}-e^{-ix})\sum_{k=0}^\infty\frac{e^{i(2k+1)x}}{2k+1}\right)\\
&=\frac14\mathrm{Re}\left(2+i\sin(x)\log\left(\frac{1+e^{ix}}{1-e^{ix}}\right)\right)\\
&=\frac14\mathrm{Re}\left(2+i\sin(x)\log\left(i\tan(x/2)\right)\right)\\
&=\frac14\mathrm{Re}\left(2-\frac\pi2\sin(x)+i\sin(x)\log(\tan(x/2))\right)\\
&=\frac12-\frac\pi8\sin(x)
\end{align}
$$
Plugging in $x=1$ yields
$$
\sum_{k=1}^\infty\frac{\cos^2(k)}{4k^2-1}=\frac12-\frac\pi8\sin(1)
$$
A: Express the sum as follows:
$$\sum_{k=1}^{\infty} \frac{\cos^2{k}}{4 k^2-1} = \frac12 \sum_{k=1}^{\infty} \frac{1}{4 k^2-1} +  \frac12 \sum_{k=1}^{\infty} \frac{\cos{2 k}}{4 k^2-1}$$
Now, the first sum on the RHS is equal to $1/2$.  This may be shown using residue theory very easily:
$$\sum_{k=-\infty}^{\infty} \frac{1}{4 k^2-1} = -\text{Res}_{z=\pm 1/2} \frac{\pi \cot{\pi z}}{4 z^2-1} = 0$$
which means that
$$2 \sum_{k=1}^{\infty} \frac{1}{4 k^2-1} - 1 = 0$$
For the other sum, break into partial fractions and reorganize.  You end up with
$$\begin{align}\sum_{k=1}^{\infty} \frac{\cos{2 k}}{4 k^2-1} &= \cos{2} +  \sum_{k=1}^{\infty} \frac{\cos{2 (k+1)}-\cos{2 k}}{2 k+1}\\ &= \cos{2} - \sin{1} \sum_{k=1}^{\infty} \frac{\sin{(2 k+1)}}{2 k+1}\end{align}$$
That last sum may be derived from the well-known sum
$$\sum_{k=0}^{\infty} \frac{\sin{(2 k+1)}}{2 k+1} = \frac{\pi}{4}$$
Therefore, I get as the sum
$$\begin{align}\sum_{k=1}^{\infty} \frac{\cos^2{k}}{4 k^2-1} &= \frac14 + \frac14 \left (\cos{2} - \frac{\pi}{4}\sin{1} + 2 \sin^2{1} \right )\\ &= \frac{1}{2} - \frac{\pi}{8} \sin{1}\end{align}$$
You may verify that this checks out numerically in Mathematica or WA.
ADDENDUM
The "well-known" sum may be evaluated by considering the series expansion of the function $\text{arctanh}{z}$:
$$\text{arctanh}{z} = \sum_{k=0}^{\infty} \frac{z^{2 k+1}}{2 k+1}$$
so that
$$
\begin{align}\sum_{k=0}^{\infty} \frac{\sin{(2 k+1)}}{2 k+1} &= \Im{\left [\sum_{k=0}^{\infty} \frac{e^{i(2 k+1)}}{2 k+1}\right]}\\ &= \frac12 \Im{\left[\text{arctanh}{e^i}\right]}\\ &= \frac12 \Im{\left[\log{\left (\frac{1+e^i}{1-e^i} \right )}\right]}\\ &= \frac12 \Im{\left[\log{\left (i \frac{\cos{1/2}}{\sin{1/2}} \right )}\right]}\\ &= \frac{\pi}{4} \end{align}$$
