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What options do I have for this series? No idea how to do it.

$$\sum_{k=1}^{\infty} \frac{\cos^2(k)}{4k^2-1}$$

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Are you sure the sum is written correctly? In particular, does it really have $\cos(k)$ in the numerator or should it be something like $\cos(k\pi)$? If it is correct, I don't think there is any method for evaluating the sum exactly. –  Peter Košinár May 27 '13 at 13:15
    
@PeterKošinár yes, it's absolutely correct. Moreover, I expect it has a nice closed form (as I was told). –  Crazy_bloke May 27 '13 at 13:17
    
@Crazy_bloke: please remember to accept an answer if it was useful to you. –  Ron Gordon May 29 '13 at 23:35

2 Answers 2

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 $$ \sum_{k=1}^\infty\frac{\cos^2(k)}{4k^2-1}=\frac12-\frac\pi8\sin(1) $$

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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

$$ \begin{align}\sum_{k=0}^{\infty} \frac{\sin{(2 k+1)}}{2 k+1} &= \Im{\left [\sum_{k=0}^{\infty} \frac{e^{i(2 k+1)}}{2 k+1}\right]}\\ &= \frac12 \Im{\left[\text{arctanh}{e^i}\right]}\\ &= \frac12 \Im{\left[\log{\left (\frac{1+e^i}{1-e^i} \right )}\right]}\\ &= \frac12 \Im{\left[\log{\left (i \frac{\cos{1/2}}{\sin{1/2}} \right )}\right]}\\ &= \frac{\pi}{4} \end{align}$$

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For the first term, no residues needed. The same partial fractions decomposition gives a telescoping series immediately. I guess the "well-known sum" can be done by summation by parts, too. –  Ted Shifrin May 27 '13 at 15:19
    
@TedShifrin: I am very residue-oriented, so that's the method that came to mind. Of course there are other ways. –  Ron Gordon May 27 '13 at 15:24
    
I too love residues, but I try to keep in mind the possible levels of the OP, so I just wanted to add the alternative solution. –  Ted Shifrin May 27 '13 at 15:32

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