# Finding $\sum_{n=1}^{\infty }\frac{243}{16(n\pi )^5}\sin(2n\pi /3)$

The WolfarmAlpha couldn't give me the sum of $$\sum_{n=1}^{\infty }\frac{243}{16(n\pi )^5}\sin(2n\pi /3)$$ therefore I thought that this problem is difficult so I used my calculator to get $(1/24)$

Is this value right or not? If this right, why the WolfarmAlph couldn't find it?

• Wait, your calculator could sum this up whist WolframAlpha could not?! What brand of calculator are you using? Dec 3, 2014 at 19:55
• Casio Super-FX PLUS $5800P$ Dec 3, 2014 at 19:57
• Trivial for Srinivasa Ramanujan :P Dec 3, 2014 at 20:25
• Factor all constants outside of the actual sum, and break the series into three subseries, then use the known values for the Dirichlet $\beta$ function. Dec 3, 2014 at 20:30
• Why the heck are all the current answers getting a downvote? Dec 4, 2014 at 2:22

Consider the sum:

$$\sum_{n=1}^{\infty} \frac{\displaystyle\sin{\frac{2 n \pi}{3}}}{n^5}$$

This sum is a bit easier than it looks, if you know the residue theorem. The main observation is that the numerator either takes the value $\sqrt{3}/2$, $-\sqrt{3}/2$, or $0$. Then you may rewrite the sum as

$$\frac{\sqrt{3}}{2} \sum_{n=1}^{\infty} \left [\frac1{(3 n-2)^5} - \frac1{(3 n-1)^5}\right ] = \frac{\sqrt{3}}{4 \cdot 3^5} \sum_{n=-\infty}^{\infty} \left [\frac1{(n-2/3)^5} - \frac1{(n-1/3)^5}\right ]$$

We may apply the residue theorem to this sum by considering the following result:

$$\sum_{n=-\infty}^{\infty} f(n) = -\pi \sum_k \operatorname*{Res}_{z=z_k} [ \cot{\pi z} \, f(z) ]$$

where $z_k$ are the non-integer poles of $f$. All we need to do is compute the residues at the poles, which in this case are at $z=1/3$ and $z=2/3$. For example,

\begin{align}\operatorname*{Res}_{z=2/3} \left [ \cot{\pi z} \, \left ((z-2/3)^{-5}-(z-1/3)^{-5} \right ) \right] &= \frac1{4!} \left [\frac{d^4}{dz^4} \cot{\pi z} \right ]_{z=2/3} - \frac1{4!} \left [\frac{d^4}{dz^4} \cot{\pi z} \right ]_{z=1/3} \\ &= -\frac{8\pi^4}{3 \sqrt{3}}\end{align}

The sum is then

$$\sum_{n=1}^{\infty} \frac{\displaystyle\sin{\frac{2 n \pi}{3}}}{n^5} = \frac{2 \pi^5}{729}$$

The result follows.

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large% \sum_{n\ =\ 1}^{\infty}{243 \over 16\pars{n\pi}^{5}}\,\sin\pars{2n\pi \over 3}} ={243 \over 16\pi^{5}}\,\Im\sum_{n\ =\ 1}^{\infty} {\pars{\expo{2\pi\ic/3}}^{n} \over n^{5}} ={243 \over 16\pi^{5}}\,\color{#c00000}{\Im\Li{5}\pars{\expo{2\pi\ic/3}}} \end{align} where $\ds{\Li{s}\pars{z}}$ is the PolyLogarithm Function .

With Jonquiere Inversion Formula $\ds{\Li{n}\pars{\expo{2\pi\ic x}} + \pars{-1}^{n}\Li{n}\pars{\expo{-2\pi\ic x}} =-\,{\pars{2\pi\ic}^{n} \over n!}\,\,{\rm B}_{n}\pars{x}}$ where $\ds{\,{\rm B}_{n}\pars{x}}$ is a Bernoulli Polynomial we'll have: \begin{align} \color{#c00000}{\Im\Li{5}\pars{\expo{2\pi\ic/3}}} &={\Li{5}\pars{\expo{2\pi\ic/3}} - \Li{5}\pars{\expo{-2\pi\ic/3}} \over 2\ic} =-\pi\,{\pars{2\pi\ic}^{4} \over 5!}\,\,{\rm B}_{5}\pars{1 \over 3} \\[5mm]&=\color{#c00000}{-\,{2\pi^{5} \over 15}\,\,{\rm B}_{5}\pars{1 \over 3}} \end{align} Note that $\ds{\,{\rm B}_{5}\pars{x}= x^{5} - {5 \over 2}\,x^{4} + {5 \over 3}\,x^{3} - {1 \over 6}\,x}$ such that $\ds{\color{#c00000}{\,{\rm B}_{5}\pars{1 \over 3} = -\,{5 \over 243}}}$.

\begin{align}&\color{#66f}{\large% \sum_{n\ =\ 1}^{\infty}{243 \over 16\pars{n\pi}^{5}}\,\sin\pars{2n\pi \over 3}} ={243 \over 16\pi^{5}}\bracks{\pars{-\,{2\pi^{5} \over 15}}\pars{-\,{5 \over 243}}} =\color{#66f}{\large{1 \over 24}} \end{align}

We have the following identity for Polylogarithm function: $$\text{Li}_n(e^{2 \pi i x}) + (-1)^n\text{Li}_n(e^{-2 \pi i x}) = - \dfrac{(2 \pi i)^n}{n!} B_n(x)$$ where $B_n(x)$ are the Bernoulli polynomials. In your case, it is easy to show that $$\sum_{n=1}^{\infty} \dfrac{\sin(nx)}{n^5} = \dfrac1{2i} \left(\text{Li}_5(e^{ix}) - \text{Li}_5(e^{-ix})\right) = -\dfrac1{2i} \dfrac{(2 \pi i)^5}{5!} B_5\left(\frac{x}{2\pi} \right) = -\dfrac{2 \pi^5}{15}B_5\left(\dfrac{x}{2\pi} \right)$$ Plugging in $x = \frac{2 \pi}3$, we obtain $$\sum_{n=1}^{\infty} \dfrac{\sin(2n\pi/3)}{n^5} = -\dfrac{2 \pi^5}{15}B_5\left(\frac13 \right) = -\dfrac{2 \pi^5}{15} \cdot \dfrac{-5}{243} = \dfrac{2 \pi^5}{729}$$

WolframAlpha thinks that the answer can be written as polylogarithm functions:

It also equals to 1/24:

• how you know this equal to $1/24$ Dec 3, 2014 at 20:18
• The first result is just the definition of $Li_n(z)$; it's the fact that it's equal to $1/24$ that's the nontrivial part. Dec 3, 2014 at 20:33

In Maple 18:

S:= Sum(243/16/(n*Pi)^5*sin(2*n*Pi/3),n=1..infinity):
simplify(value(convert(S,exp)));


$$\dfrac{1}{24}$$

• This should have been a comment! Dec 4, 2014 at 1:25
• @Integrator I disagree. This answer is indeed answering the question as it is written: "Is this value right or not?" Dec 4, 2014 at 5:01
• Sorry @Venus didn't read! +1 For that! Dec 4, 2014 at 5:03