# Calculate $\sum \limits_{n = 1}^{\infty} \frac{\cos 2n}{n^2}$

I need to calculate the following sum:

$$\sum \limits_{n = 1}^{\infty} \frac{\cos 2n}{n^2}$$

I've tried adding an imaginary part and differentiating:

$$f(x) = \sum \limits_{k = 1}^{\infty} \frac{\cos 2xk + i \sin 2xk}{k^2} \\ f(x) = \sum \limits_{k = 1}^{\infty} \frac{e^{2ixk}}{k^2} \\ f'(x) = \sum \limits_{k = 1}^{\infty} \frac{2i e^{2ixk}}{k} \\ f''(x) = - 4\sum \limits_{k = 1}^{\infty} e^{2ixk} \\ f''(x) = -4\frac{e^{2ix}}{1 - e^{2ix}}$$

Where $$f(x)$$ if a function of which I need to find the value at $$x = 1$$.

After differentiating once I get

$$f'(x) = \frac{\log \left( 1 - e^{kx} \right)}{k} + C$$

(k is just some constant), and I can't integrate once more as I'll get an integral logarithm which I don't want to work with.

Is there any more pleasant way to calculate the aforementioned sum?

The series has terms of the form $$a_n\cos (n)$$ with $$n\in\mathbb{N}$$, $$n$$ even, so your mind should immediately jump to Fourier series, where the coefficients are of the form $$\displaystyle \frac1{n^2}$$ for even $$n$$ or $$0$$ otherwise. Insert a variable into the series to transform it into the function, $$\displaystyle \sum_{n=1}^{\infty}\frac{\cos\left(2nx\right)}{n^{2}}$$. We see from a graph that this is a $$\pi$$-periodic, U-shaped curve with a minimum at $$x=\frac\pi2$$ so we can make the ansatz that it is the Fourier series of parabola of the form $$\left(x-\frac{\pi}{2}\right)^2$$ up to a constant difference. We can then treat this with a typical Fourier series of the function over $$(-\pi,\pi)$$ by taking the absolute value of $$x$$.

The function is even so the coefficients of $$\sin nx$$ terms, $$b_n$$, are all $$0$$. We may then solve for $$\displaystyle a_0=\frac1\pi\int_\pi^\pi\left(|x|-\frac\pi2\right)^2\,\mathrm{d}x=\frac{\pi^{2}}{6}$$ and, using $$\sin(\pi n)=0$$ and $$\cos(\pi n)=(-1)^n$$,

\begin{align} a_n&=\frac{1}{\pi}\int_{-\pi}^{\pi}\left(\left|x\right|-\frac{\pi}{2}\right)^{2}\cos\left(nx\right)\,\mathrm{d}x \\ &=\frac{2}{\pi}\cdot\frac{\left(\pi^{2}-8\right)\sin\left(\pi n\right)+4\pi n+4\pi n\cos\left(\pi n\right)}{4n^{3}} \\ &=2\cdot \frac{1+\left(-1\right)^{n}}{n^{2}} \end{align}

Thus, $$\displaystyle \left(x-\frac\pi2\right)^2=\frac{\pi^{2}}{12}+2\sum_{n=1}^{\infty}\frac{1+\left(-1\right)^{n}}{n^{2}}\cos\left(nx\right)$$. You should be able to take it from here.

$$\sum_{k=1}^\infty \frac{\cos 2 k x}{k^2} \sim f(x) = \left(x-\frac{\pi}{2}\right)^2 - \frac{\pi^2}{12} .$$

See, for example, this math stackexchange exchange.

This means that

$$\sum_{k=1}^\infty \frac{\cos 2 k }{k^2} =\left(1-\frac{\pi}{2}\right)^2 - \frac{\pi^2}{12},$$

since the series converges to $$f(x)$$ on $$[0,\pi]$$ and its periodic extension everywhere.

Subtract $$\sum_n\frac1{n^2}=\frac{\pi^2}6$$ to obtain

$$\begin{eqnarray} \sum_{n=1}^\infty\frac{\cos2n}{n^2}-\frac{\pi^2}6 &=& \sum_{n=1}^\infty\frac{1-\cos2n}{n^2} \\ &=& -2\sum_{n=1}^\infty\left(\frac{\sin n}n\right)^2\;. \end{eqnarray}$$

Now note that $$\frac1\pi\frac{\sin n}n$$ is the $$n$$-th coefficient in the Fourier series of a rectangular pulse with period $$2\pi$$ and length $$2$$:

$$\frac1{2\pi}\int_{-1}^1\mathrm e^{-\mathrm inx}\mathrm dx=\frac1\pi\frac{\sin nx}n\;.$$

By Parseval’s theorem we have

$$\sum_{n=-\infty}^\infty\left(\frac1\pi\frac{\sin nx}n\right)^2=\frac1{2\pi}\int_{-1}^1\mathrm dx=\frac1\pi\;,$$

and thus

$$\begin{eqnarray} \sum_{n=1}^\infty\left(\frac{\sin n}n\right)^2 &=& \frac12\left(\sum_{n=-\infty}^\infty\left(\frac{\sin nx}n\right)^2-1\right) \\ &=& \frac{\pi-1}2\;. \end{eqnarray}$$

Substituting above yields

$$\begin{eqnarray} \sum_{n=1}^\infty\frac{\cos2n}{n^2} &=& \frac{\pi^2}6-2\cdot\frac{\pi-1}2 \\ &=&\frac{\pi^2}6-\pi+1\;. \end{eqnarray}$$