Proving $$\sum_{n=1}^{\infty }\frac{\cos(n)}{n^4}=\frac{\pi ^4}{90}-\frac{\pi ^2}{12}+\frac{\pi }{12}-\frac{1}{48}$$

I tried with Wolfram but it couldn't give me any clear value as shown below enter image description here

The numerical value of Wolfram not different of my closed-form. Can anyone explain how the $.5(Li_4(e^{-i}+Li_4(e^{i}))$ equal the above closed-form

  • 1
    $\begingroup$ Wolfram Alpha's answer is essentially trivial. $$\mathrm{Li}_4(z)=\sum_{n=1}^{\infty} \frac{z^n}{n^4}$$ by definition. And WA is saying that your sum is the real part of $\mathrm{Li}_4(e^i)$. $\endgroup$ Mar 28, 2015 at 22:14
  • $\begingroup$ @ThomasAndrews, your explaination is ok , now, how can I prove the closed form depending on what you posted? $\endgroup$
    – E.H.E
    Mar 28, 2015 at 22:20
  • $\begingroup$ I don't know, otherwise, I would have posted an answer. The question is not trivial, but Wolfram's answer is. :) $\endgroup$ Mar 28, 2015 at 22:21
  • 1
    $\begingroup$ This question is a duplicate. I found an answer among my posts at this MSE link II. $\endgroup$ Mar 28, 2015 at 22:41
  • 1
    $\begingroup$ @Marko Riedel Interesting. I myself was hoping for a more elegant (and less dependent on advanced, non-trivial theorems) solution, which might still exist for this particular case. Therefore I don't feel like we should close the question just yet. $\endgroup$ Mar 28, 2015 at 22:49

4 Answers 4


Start with : $$\tag{1}f(x):=\sum_{n=1}^{\infty }\frac{\cos(n\,x)}{n^4}$$ and observe that (from the "Sawtooth Wave" Fourier series with $L=2\pi$ and $x=2X$) : $$\tag{2}f^{(3)}(x)=\sum_{n=1}^{\infty }\frac{\sin(n\,x)}{n}=\frac{\pi-x}2,\quad \text{for}\;x\in(0,2\pi)$$

At this point it remains only to integrate $(2)$ three times with the appropriate constant of integration :

$$\tag{3}f^{(2)}(x)=\sum_{n=1}^{\infty }\frac{-\cos(n\,x)}{n^2}=C+\frac{\pi}2x-\frac 14x^2$$ with $\;\displaystyle C=f^{(2)}(0)=-\zeta(2)=-\frac{\pi^2}6$. $$\tag{4}f'(x)=\sum_{n=1}^{\infty }\frac{-\sin(n\,x)}{n^3}=-\frac{\pi^2}6x+\frac{\pi}4x^2-\frac 1{12}x^3$$ (since $f'(0)=0$) and the final solution : $$\tag{5}f(x)=\sum_{n=1}^{\infty }\frac{\cos(n\,x)}{n^4}=\zeta(4)-\frac{\pi^2}{12}x^2+\frac{\pi}{12}x^3-\frac 1{48}x^4,\quad \text{for}\;x\in(0,2\pi)$$ Of course $x=1$ and $\zeta(4)=\dfrac{\pi^4}{90}$ will return the wished conclusion.


It is clear that $\;\displaystyle\sum_{n=1}^{\infty }\frac{\cos(n\,x)}{n^{2m}}\;$ and $\;\displaystyle\sum_{n=1}^{\infty }\frac{\sin(n\,x)}{n^{2m-1}}\;$ may be evaluated by this method for $x\in(0;2\pi)$ and $m$ any positive integer. The polynomials obtained are well known since they correspond to the Bernoulli polynomials as given in Abramowitz and Stegun $(23.1.18)$ and $(23.1.17)$.

The remaining cases $\;\displaystyle\sum_{n=1}^{\infty }\frac{\cos(n\,x)}{n^{2m-1}}\;$ and $\;\displaystyle\sum_{n=1}^{\infty }\frac{\sin(n\,x)}{n^{2m}}\;$ are not so easy to evaluate and got the name of Clausen functions $\;\operatorname{Cl}_{2m-1}(x)\;$ and $\;\operatorname{Cl}_{2m}(x)$.

The origin of the difficulty is that $\;\displaystyle\operatorname{Cl}_1(x):=\sum_{n=1}^{\infty }\frac{\cos(n\,x)}{n}=-\ln(2\,\sin(x/2))\;$ (see for example here) implying that the integrals will be harder to evaluate (except for specific fractions of $\pi$).


$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\sum_{n = 1}^{\infty} {\cos\pars{n} \over n^{4}}} = \Re\sum_{n = 1}^{\infty} {\pars{\expo{\ic}}^{n} \over n^{4}} = \Re\on{Li}_{4}\pars{\expo{\ic}} \\[5mm] = & \ {1 \over 2}\left[% \on{Li}_{4}\pars{\exp\pars{2\pi\ic{1 \over 2\pi}}}\right. \\[2mm] & \left. \phantom{AA}+ \pars{-1}^{4}\, \on{Li}_{4}\pars{% \exp\pars{-2\pi\ic{1 \over 2\pi}}} \right] \\[5mm] = & \ {1 \over 2}\bracks{-\,{\pars{2\pi\ic}^{4} \over 4!}\on{B}_{4}\pars{1 \over 2\pi}}\tag{1}\label{1} \end{align} $\ds{\on{Li}_{s}\ \mbox{and}\ \on{B}_{n}}$ are the Polylogarithm Function and a Bernoulli Polynomial, respectively.

The bracket enclosed identity in (\ref{1}) is the $\ds{Jonqui\grave{e}re\ Inversion\ Formula}$. $$ \mbox{Note that}\quad \on{B}_{4}\pars{x} = x^{4} - 2x^{3} + x^{2} - {1 \over 30} $$

Therefore, with (\ref{1}), \begin{align} & \color{#44f}{\sum_{n = 1}^{\infty} {\cos\pars{n} \over n^{4}}} \\[5mm] = & \ {1 \over 2}\braces{-\,{16\pi^{4} \over 24} \bracks{\pars{1 \over 2\pi}^{4} - 2\pars{1 \over 2\pi}^{3} + \pars{1 \over 2\pi}^{2} - {1 \over 30}}} \\[5mm] = & \ \bbx{\color{#44f}{{\pi^{4} \over 90}- {\pi^{2} \over 12} + {\pi \over 12} - {1 \over 48}}} \approx 0.5008 \\ & \end{align}


By using the series \begin{align} \sum_{n=1}^{\infty} \frac{ \cos(nx) }{ n^{2} } = \zeta(2) - \frac{ \pi \, x}{2} + \frac{x^{2}}{4} \end{align} integrate with respect to $x$ from zero to $t$ to obtain \begin{align} \sum_{n=1}^{\infty} \frac{\sin(n t)}{n^{3}} = \zeta(2) \, t - \frac{\pi \, t^{2}}{4} + \frac{t^{3}}{12}. \end{align} Integrate once again in a similar manor to obtain \begin{align} \sum_{n=1}^{\infty} \frac{\cos(nx)}{n^{4}} = \zeta(4) - \frac{\zeta(2) \, x^{2}}{2} + \frac{\pi \, t^{3}}{12} - \frac{x^{4}}{48}. \end{align} Upon letting $x=1$ the above results yield \begin{align} \sum_{n=1}^{\infty} \frac{\cos(n)}{n^{2}} &= \zeta(2) + \frac{1 - 2 \pi}{4} \\ \sum_{n=1}^{\infty} \frac{\sin(n)}{n^{3}} &= \zeta(2) + \frac{1 - 3 \pi}{12} \\ \sum_{n=1}^{\infty} \frac{\cos(n)}{n^{4}} &= \zeta(4) - \frac{\zeta(2)}{2} + \frac{\pi}{12} - \frac{1}{48}. \end{align}

  • $\begingroup$ I think there is a problem with the signs of terms $\endgroup$
    – E.H.E
    Mar 28, 2015 at 23:19
  • $\begingroup$ @Ehegh Thanks for the correction. $\endgroup$
    – Leucippus
    Mar 29, 2015 at 0:25

Let $k$ be a nonnegative integer.

Contour integration shows that $$ \sum_{n=1}^{\infty}\frac{\sin (n)}{n^{2k-1}} = - \frac{1}{2} \, \Im\, \operatorname{Res} \left[\frac{\pi \left(\cot (\pi z)-\color{red}{i} \right)e^{iz}}{z^{2k-1}},0 \right]$$ and $$\sum_{n=1}^{\infty} \frac{\cos (n)}{n^{2k}} = - \frac{1}{2} \, \Re \, \operatorname{Res} \left[\frac{\pi \left(\cot (\pi z)-\color{red}{i} \right)e^{iz}}{z^{2k}},0 \right].$$

See my previous answer here for an explanation of why we use $\cot(\pi z)-i$ and not just $\cot(\pi z)$.

Basically it has to do with the fact that the magnitude of $e^{iz}$ is unbounded in the lower half-plane.

At the origin, we have the Laurent series expansion $$ \begin{align} \pi \left(\cot(\pi z)-i \right)e^{iz} &=\small \left(\frac{1}{z}- i \pi - 2\zeta(2)z -2 \zeta(4) x^{3} + O(z^{5}) \right) \left( 1+iz- \frac{z^{2}}{2!}- \frac{iz^{3}}{3!} + \frac{z^{4}}{4!} + O(z^{5}) \right) \\ & = \small \frac{1}{z} + i \left(1-\pi \right) + \left(- \frac{1}{2!} +\pi - 2\zeta(2) \right)z + i \left(- \frac{1}{3!} + \frac{\pi}{2!} -2 \zeta(2) \right)z^{2} \\ &+ \small \left(\frac{1}{4!} -\frac{\pi}{3!} + \frac{2 \zeta(2)}{2!} - 2 \zeta(4) \right)z^{3} + O(z^{4}) . \end{align}$$

Therefore, $$\sum_{n=1}^{\infty} \frac{\sin (n)}{n} = \frac{\pi -1}{2}, $$

$$\sum_{n=1}^{\infty} \frac{\cos (n)}{n^{2}} = \frac{1}{4} - \frac{\pi}{2} + \zeta(2), $$

$$\sum_{n=1}^{\infty} \frac{\sin (n)}{n^{3}} = \frac{1}{12} - \frac{\pi}{4} + \zeta(2), $$

$$\sum_{n=1}^{\infty} \frac{\cos (n)}{n^{4}} = -\frac{1}{48} + \frac{\pi}{12} - \frac{\zeta(2)}{2} + \zeta(4), $$ and so on.


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