# Show that $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}}\cos nx=\frac{1}{12}(3x^{2}-\pi^2)$ for $-\pi \leq x \leq \pi$

How do we show that for $$-\pi \leq x \leq \pi$$, $$\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}}\cos nx=\frac{1}{12}(3x^{2}-\pi^2)$$

I know that without $$(-1)^{n}$$ term, the series converges to $$\frac{x^{2}}{4}-\frac{\pi x }{2}+\frac{\pi^{2}}{6}$$ for $$x$$ in $$[0,2\pi]$$. But I'm not sure how to find the sum if there is a $$(-1)^{n}$$ term there.

• @PeterForeman that's true, but I guess the major problem is determine rhs by lhs Jul 27, 2020 at 18:47
• Sometimes it works: let $S$ be rhs. Now differentiate it. Maybe its easier to check derivative. Or second derivative. Jul 27, 2020 at 18:51
• Hint: Replace $x$ by $x-\pi$ in the LHS and use the addition formulae. You should immediately see the form of your answer without the $(-1)^n$. Also the domain changes correspondingly. Jul 27, 2020 at 18:52
• split the sum into even /odd n's Jul 27, 2020 at 19:00

You could use the Fourier Series of $$f(x)=\frac1{12}(3x^2-\pi^2)$$ in $$[-\pi,\pi].$$ $$f(x)=\frac{a_0}2+\sum_{n=1}^\infty[a_n\cos(nx)+b_n\sin(nx)]$$where$$a_{n\ge0}=\frac1\pi\int_{-\pi}^\pi f(x)\cos(nx)~dx=\frac1{12\pi}\int_{-\pi}^\pi(3x^2-\pi^2)\cos(nx)~dx$$For $$n=0$$,$$a_0=\frac1{6\pi}[x^3-\pi^2x]_0^\pi=0$$For $$n\ge1$$,

$$a_n=\frac1{12\pi}\left\{\left[(3x^2-\pi^2)\int\cos(nx)~dx\right]_{-\pi}^\pi-6\int_{-\pi}^\pi x\int\cos(nx)~dx~dx\right\}\\=-\frac1{n\pi}\int_0^\pi x\sin(nx)~dx=-\frac1{n\pi}\left\{\left[x\int\sin(nx)~dx\right]^\pi_0-\int_0^\pi\int\sin(nx)~dx\right\}\\=\frac{\cos(n\pi)}{n^2}$$

and$$b_{n\ge1}=\frac1\pi\int_{-\pi}^\pi f(x)\sin(nx)~dx=0~\forall n.$$ Since $$f(x)$$ is continuous in the given domain, the Fourier series$$\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\cos(nx)$$will converge to $$f(x)$$ for each point due to Dirichlet conditions.

Denote $$f(x)= \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}}\cos nx$$. Then, $$f(0)= \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}}=-\frac{\pi^2}{12}$$

\begin{align} f’(x) & = -\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}\sin nx =\frac1{2i}\sum_{n=1}^{\infty}(-1)^{n+1}\frac{e^{ixn }-e^{-ixn }}{n}\\ &=\frac1{2i}[\ln(1+e^{ix })-\ln (1+e^{-ix })] =\frac1{2i}\ln e^{ix} =\frac{x}{2}\\ \end{align}

Thus

$$\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}}\cos nx = f(0) + \int_0^x f’(t)dt = \frac{1}{12}(3x^{2}-\pi^2)$$

Let $$f(x;\lambda)$$ be given by the series

$$f(x;\lambda)=-\sum_{n=1}^\infty \lambda^n e^{inx}\tag1$$

for $$|\lambda|<1$$.

Summing the geometric series in $$(1)$$ we find that for $$|\lambda|<1$$

$$f(x;\lambda)=\frac{\lambda e^{ix}}{\lambda e^{ix}-1}\tag2$$

Exploiting the uniform convergence of $$\sum_{n=1}^\infty \lambda^n e^{inx}$$ for $$\lambda \in [-1+\delta, 1-\delta]$$, $$\delta>0$$, integrating $$f(x;\lambda)$$ in $$(1)$$, taking the real part reveals

$$\text{Re}\left(\int_0^x f(t;\lambda)\,dt\right)=-\sum_{n=1}^\infty \frac{\lambda^n\sin(nx)}{n}\tag3$$

Then, exploiting the uniform convergence of $$\sum_{n=1}^\infty \frac{\lambda^n\sin(nx)}{n}$$ for $$\lambda \in [-1,1]$$ and for $$x\in [\nu,2\pi-\nu]$$, $$\nu>0$$, we let $$\lambda \to -1^+$$ in $$(3)$$ to find

\begin{align} \lim_{\lambda\to 1^-}\text{Re}\left(\int_{0}^x f(t;\lambda)\,dt\right)&=-\sum_{n=1}^\infty \frac{(-1)^n\sin(nx)}{n}\tag4 \end{align}

Going back to $$(2)$$, we integrate $$f(x;\lambda)$$, take the real part, and let $$\lambda\to -1^+$$ to find

$$\lim_{\lambda\to -1^-}\text{Re}\left(\int_{0}^x f(t;\lambda)\,dt\right)=\frac x2\tag5$$

Integrating $$(4)$$ and $$(5)$$ once more and using $$\sum_{n=1}^\infty \frac1{n^2}=\frac{\pi^2}{12}$$ yields

$$\sum_{n=1}^\infty \frac{(-1)^n \cos(nx)}{n^2}=\frac{x^2}{4}-\frac{\pi^2}{12}$$

as was to be shown!

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ \begin{align} \left.\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^{2}}\cos\pars{nx} \,\right\vert_{\ds{\ -\pi\ \leq\ x\ \leq\ \pi}} & = \sum_{n = 1}^{\infty}{\cos\pars{n\bracks{x + \pi}} \over n^{2}} = \Re\sum_{n = 1}^{\infty}{\bracks{\expo{\ic\pars{x + \pi}}}^{n} \over n^{2}} \\[5mm] & = \Re\mrm{Li}_{2}\pars{\expo{\ic\bracks{x + \pi}}} \end{align}

where $$\ds{\mrm{Li}_{s}}$$ is a Polylogarithm.

Then, \begin{align} &\left.\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^{2}}\cos\pars{nx} \,\right\vert_{\ds{\ -\pi\ \leq\ x\ \leq\ \pi}} \\[5mm] = &\ {1 \over 2}\bracks{% \mrm{Li}_{2}\pars{\exp\pars{2\pi\,{x + \pi \over 2\pi}\ic}} + \mrm{Li}_{2}\pars{\exp\pars{-2\pi\,{x + \pi\over 2\pi}\ic}}} \\[5mm] & = {1 \over 2}\bracks{-\,{\pars{2\pi\ic}^{2} \over 2!} \,\mrm{B}_{2}\pars{x + \pi\over 2\pi}} \end{align} Here, I used Jonqui$$\grave{\mrm{e}}$$re Inversion Formula and $$\ds{\mrm{B}_{n}}$$ is a Bernoulli Polynomial. For instance, $$\ds{\mrm{B}_{2}\pars{x} = x^{2} - x + 1/6}$$.

Therefore, \begin{align} \left.\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^{2}}\cos\pars{nx} \,\right\vert_{\ds{\ -\pi\ \leq\ x\ \leq\ \pi}} & = \pi^{2}\bracks{\pars{x + \pi \over 2\pi}^{2} - {x + \pi \over 2\pi} + {1 \over 6}} \\[5mm] & = \bbx{{1 \over 12}\pars{3x^{2} - \pi^{2}}} \\[5mm] & \end{align}