Sum $\sum_{n = 1}^{\infty}\left[\frac1n\sin nx + \frac{1}{n^2}\cos nx\right]$

Question

I want to find the following sum by using the complex methods for series ($z = \cos nx + i \sin nx$). $$ \sum_{n = 1}^{\infty}\left[\frac1n\sin nx + \frac{1}{n^2}\cos nx\right] $$ Here is my attempt:

$$ S_N = \sum_{n = 1}^{N}\frac1n\sin nx + \sum_{n = 1}^{N}\frac{1}{n^2}\cos nx $$ I'm done with the first sum: $$ \sum_{n = 1}^{N}\frac1n\sin nx = \Im \sum_{n = 1}^{N}\frac{z^n}{n} \Rightarrow \lim_{N \to \infty} \sum_{n = 1}^{N}\frac1n\sin nx = -\Im \ln(1 - z) $$ But I'm stuck with the second one: $$ \sum_{n = 1}^{N}\frac{1}{n^2}\cos nx = \Re \sum_{n = 1}^{N}\frac{z^n}{n^2} $$

Answer 1

Note that we have $$\int_0^x \sin(nt)\,dt=\frac{1-\cos(nx)}{n}$$ Therefore we can write for $x\in [0,2\pi]$

$$\begin{align} \sum_{n=1}^\infty \frac{\cos(nx)}{n^2}&=\frac{\pi^2}{6}-\sum_{n=1}^\infty \frac1n \int_{0}^x \sin(nt)\,dt\\\\ &=\frac{\pi^2}{6}-\int_0^x \sum_{n=1}^\infty \frac{\sin(nt)}{n}\,dt\\\\ &=\frac{\pi^2}{6}+\int_0^x \text{Im}\left(\log(1-e^{ix})\right)\,dt\\\\ &=\frac{\pi^2}6-\int_0^x \arctan(\cot(t/2))\,dt \\\\ &=\frac{\pi^2}{6}-\int_0^x \left(\frac{\pi}{2}-\frac t2\right)\,dt\\\\ &=\frac{\pi^2}{6}-\frac\pi 2 x +\frac{x^2}{4} \end{align}$$

Answer 2

Continue with

\begin{align} \sum_{n = 1}^{\infty}\frac{\sin nx }n = -\Im \ln(1 - z) = \frac i2 [\ln( 1-z) -\ln (1-\bar z)] =\frac i2\ln(-z)= \frac{\pi-x}2 \end{align}

and use the result to evaluate

$$ \sum_{n = 1}^{\infty}\frac{1- \cos nx }{n^2}= \sum_{n = 1}^{\infty}\int_0^x \frac{\sin nt }{n}dt = \int_0^x \frac{\pi-t}{2}dt= \frac{\pi}2x-\frac14x^2 $$

Thus \begin{align} \sum_{n = 1}^{\infty}\left(\frac1n\sin nx + \frac{1}{n^2}\cos nx\right) &= \frac{\pi-x}2 - \frac{\pi}2x+\frac14x^2 + \sum_{n = 1}^{\infty}\frac{1}{n^2} \\ &=\frac{x^2}4 - \frac{(\pi+1)x-\pi}2 +\frac{\pi^2}6 \end{align}

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[15px,#ffd]{\sum_{n = 1}^{\infty} \bracks{{\sin\pars{nx} \over n} + {\cos\pars{nx} \over n^{2}}}}:\ {\large ?}}$

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\begin{align} \sum_{n = 1}^{\infty}{\sin\pars{nx} \over n} & = \Im\sum_{n = 1}^{\infty}{\pars{\expo{\ic x}}^{n} \over n} = -\,\Im\ln\pars{1 - \expo{\ic x}} \\[5mm] & = -\,\Im\ln\pars{1 - \cos\pars{x} - \ic\sin\pars{x}} \\[5mm] & = -\arctan\pars{-\sin\pars{x} \over 1 - \cos\pars{x}} = \arctan\pars{2\sin\pars{x/2}\cos\pars{x/2} \over 2\sin^{2}\pars{x/2}} \\[5mm] & = \arctan\pars{\cot\pars{x \over 2}} = \arctan\pars{\tan\pars{{\pi \over 2} - {x \over 2}}} = \bbx{{\pi \over 2} - {x \over 2}}\label{1}\tag{1} \end{align}

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\begin{align} \sum_{n = 1}^{\infty}{\cos\pars{nx} \over n^{2}} & = \Re\sum_{n = 1}^{\infty}{\pars{\expo{\ic x}}^{n} \over n^{2}} = \Re\operatorname{Li}_{2}\pars{\exp\pars{2\pi\ic\,{x \over 2\pi}}} \end{align}

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

With the Jonqui$\grave{\mrm{e}}$re Inversion Formula

\begin{align} \sum_{n = 1}^{\infty}{\cos\pars{nx} \over n^{2}} & = -\,{1 \over 2}\,{\pars{2\pi\ic}^{2} \over 2!} \operatorname{B}_{2}\pars{x \over 2\pi} \end{align}

where $\ds{\operatorname{B}_{s}}$ is a Bernoulli Polynomial. In particular, $\ds{\operatorname{B}_{2}\pars{x} = x^{2} - x + 1/6}$.

Then,

\begin{align} \sum_{n = 1}^{\infty}{\cos\pars{nx} \over n^{2}} & = \pi^{2} \bracks{\pars{x \over 2\pi}^{2} - {x \over 2\pi} + {1 \over 6}} = \bbx{{1 \over 4}\,x^{2} - {\pi \over 2}\,x + {\pi^{2} \over 6}} \label{2}\tag{2} \end{align}

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Finally, with (\ref{1}) and (\ref{2}): $$ \sum_{n = 1}^{\infty} \bracks{{\sin\pars{nx} \over n} + {\cos\pars{nx} \over n^{2}}} = \bbox[15px,#ffd,border:1px solid navy]{{1 \over 4}\,x^{2} - {1 + \pi \over 2}\,x + {\pi \over 2} + {\pi^{2} \over 6}} $$

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