1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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

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.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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}

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.