How to compute the following series:


I tried


The idea is using Riemann zeta function


but the term $2^n$ makes complicated. I know that $$\sum_{n=1}^{\infty}\frac{1}{2^n}=1$$ using geometric series but I don't know how to use those series to answer the question. Any help would be appreciated. Thanks in advance.

  • 2
    $\begingroup$ Have you tried integrating $1+x+x^2+\cdots=1/(1-x)$? $\endgroup$ – abnry May 26 '14 at 22:19
  • $\begingroup$ @nayrb Not yet but how to use that to answer the question? I still don't get it $\endgroup$ – Venus May 26 '14 at 22:28
  • $\begingroup$ This is a polylogarithm. $\endgroup$ – Lucian May 27 '14 at 2:21

Consider Maclaurin series of natural logarithm $$ \ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}. $$ Taking $x=\dfrac12$ yields \begin{align} \ln\left(1-\frac12\right)&=-\sum_{n=1}^\infty\frac{1}{2^n\ n}\\ \ln2&=\sum_{n=1}^\infty\frac{1}{2^n\ n}. \end{align} Now, dividing the Maclaurin series of natural logarithm by $x$ yields \begin{align} \frac{\ln(1-x)}{x}&=-\sum_{n=1}^\infty\frac{x^{n-1}}{n}, \end{align} then integrating both sides and taking the limit of integration $0<x<\dfrac12$. We obtain \begin{align} \int_0^{\Large\frac12}\frac{\ln(1-x)}{x}\ dx&=-\int_0^{\Large\frac12}\sum_{n=1}^\infty\frac{x^{n-1}}{n}\ dx\\ &=-\sum_{n=1}^\infty\int_0^{\Large\frac12}\frac{x^{n-1}}{n}\ dx\\ &=-\left.\sum_{n=1}^\infty\frac{x^{n}}{n^2}\right|_{x=0}^{\Large\frac12}\\ -\int_0^{\Large\frac12}\frac{\ln(1-x)}{x}\ dx&=\sum_{n=1}^\infty\frac{1}{2^n\ n^2}. \end{align} The integral in the LHS is $\text{Li}_2\left(\dfrac12\right)=\dfrac{\pi^2}{12}-\dfrac{\ln^22}{2}$, but since you are not familiar with dilogarithm function, we will evaluate the LHS integral using IBP. Taking $u=\ln(1-x)$ and $dv=\dfrac1x\ dx$, we obtain \begin{align} \int_0^{\Large\frac12}\frac{\ln(1-x)}{x}\ dx&=\left.\ln(1-x)\ln x\right|_0^{\large\frac12}+\int_0^{\Large\frac12}\frac{\ln x}{1-x}\ dx\\ &=\ln^22-\lim_{x\to0}\ln(1-x)\ln x-\int_1^{\Large\frac12}\frac{\ln (1-x)}{x}\ dx\ ;\\&\color{red}{\Rightarrow\text{let}\quad x=1-x}\\ \int_0^{\Large\frac12}\frac{\ln(1-x)}{x}\ dx+\int_1^{\Large\frac12}\frac{\ln (1-x)}{x}\ dx&=\ln^22-0\\ -\left.\sum_{n=1}^\infty\frac{x^{n}}{n^2}\right|_{x=0}^{\Large\frac12}-\left.\sum_{n=1}^\infty\frac{x^{n}}{n^2}\right|_{x=1}^{\Large\frac12}&=\ln^22\\ \sum_{n=1}^\infty\frac{1}{2^n\ n^2}+\sum_{n=1}^\infty\frac{1}{2^n\ n^2}-\sum_{n=1}^\infty\frac{1}{n^2}&=-\ln^22\\ 2\sum_{n=1}^\infty\frac{1}{2^n\ n^2}-\frac{\pi^2}{6}&=-\ln^22\\ \sum_{n=1}^\infty\frac{1}{2^n\ n^2}&=\frac{\pi^2}{12}-\frac{\ln^22}{2}. \end{align} Thus, \begin{align} \sum_{n=1}^\infty\frac{n+1}{2^n\ n^2}&=\sum_{n=1}^\infty\left(\frac{1}{2^n\ n}+\frac{1}{2^n\ n^2}\right)\\ &=\large\color{blue}{\ln2+\frac{\pi^2}{12}-\frac{\ln^22}{2}}. \end{align}

  • 5
    $\begingroup$ Nice explanation. +1 $\endgroup$ – Random Variable May 27 '14 at 1:24
  • 1
    $\begingroup$ @RandomVariable Yup! Very nice. I like it your answer Tunk-Fey. Thanks! You help me twice. IOU big! $\endgroup$ – Venus May 27 '14 at 1:28
  • $\begingroup$ @RandomVariable Thanks Mr. Feynman, I'm honored. :) $\endgroup$ – Tunk-Fey May 27 '14 at 1:54
  • 4
    $\begingroup$ @Tunk-Fey Good work, very nice man! A young mathematician on the rise... +1 $\endgroup$ – Jeff Faraci May 27 '14 at 2:52
  • $\begingroup$ @Integrals Thanks Jeff. I'm flattered. :) $\endgroup$ – Tunk-Fey May 27 '14 at 3:50

The series \begin{align} S = \sum_{n=1}^{\infty} \frac{n+1}{2^{n} \ n^{2}} \end{align} can be expressed as \begin{align} S = \sum_{n=1}^{\infty} \frac{1}{2^{n} \ n} + \sum_{n=1}^{\infty} \frac{1}{2^{n} \ n^{2}} \end{align} and is seen to be \begin{align} S = - \ln\left( 1 - \frac{1}{2} \right) + Li_{2}\left(\frac{1}{2}\right), \end{align} where $Li_{2}(x)$ is the dilogarithm function. Since \begin{align} Li_{2}\left(\frac{1}{2}\right) = \frac{\pi^{2}}{12} - \frac{1}{2} \ \ln^{2}(2) \end{align} then the resulting series has the value \begin{align} \sum_{n=1}^{\infty} \frac{n+1}{2^{n} \ n^{2}} = \frac{\pi^{2}}{12} + \ln(2) - \frac{1}{2} \ \ln^{2}(2). \end{align} This may also be seen in the form \begin{align} \sum_{n=1}^{\infty} \frac{n+1}{2^{n} \ n^{2}} = \frac{\pi^{2}}{12} + \frac{1}{2} \ \ln(2) \ \ln\left(\frac{e^{2}}{2}\right). \end{align}

  • $\begingroup$ What is the dilogarithm function? I don't learn that subject yet. I hope you provide me a conventional method to answer this question. BTW, thank for your answer. +1 $\endgroup$ – Venus May 26 '14 at 23:11
  • $\begingroup$ @Venus If you like an answer you can hit the check mark to accept the answer. Leucippus very clever solution, nice! $\endgroup$ – Jeff Faraci May 26 '14 at 23:12
  • 1
    $\begingroup$ @Integrals I know that but I haven't found a convenience answer yet $\endgroup$ – Venus May 26 '14 at 23:16
  • 1
    $\begingroup$ A start for the dilogarithm function is the Mathworld page mathworld.wolfram.com/Dilogarithm.html . It has a connection to the Hurwitz Zeta which is a general form of the zeta function you have listed in the statement of the problem. $\endgroup$ – Leucippus May 26 '14 at 23:17
  • 1
    $\begingroup$ This is the solution. Dilog function is here en.wikipedia.org/wiki/Dilogarithm, this is the most convenient answer you will get my friend, it is quite common dilog.. @Venus. $\endgroup$ – Jeff Faraci May 26 '14 at 23:17

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\sum_{n = 1}^{\infty}{n + 1 \over 2^{n}\,n^{2}}:\ {\large ?}}$

\begin{align}&\color{#c00000}{% \sum_{n = 1}^{\infty}{n + 1 \over 2^{n}\,n^{2}}} =\sum_{n = 1}^{\infty}{\pars{1/2}^{n} \over n} +\sum_{n = 1}^{\infty}{\pars{1/2}^{n} \over n^{2}} ={\rm Li}_{1}\pars{\half} + {\rm Li}_{2}\pars{\half} \end{align} where $\ds{{\rm Li_{s}}\pars{z} \equiv \sum_{k = 1}^{\infty}{z^{k} \over k^{\rm s}}}$ is the PolyLogarithm Function.

However: $$ {\rm Li}_{1}\pars{\half} = \ln\pars{2}\qquad\mbox{and}\qquad {\rm Li}_{2}\pars{\half} = {\pi^{2} \over 12} - \half\,\ln^{2}\pars{2} $$

$$\color{#00f}{\large% \sum_{n = 1}^{\infty}{n + 1 \over 2^{n}\,n^{2}} ={\pi^{2} \over 12} + \ln\pars{2} - \half\,\ln^{2}\pars{2}}\approx 1.2754 $$

  • $\begingroup$ If I use dilog function maybe my answer would be shorter, but unfortunately the OP doesn't understand dilog function. $\endgroup$ – Tunk-Fey May 27 '14 at 1:22
  • $\begingroup$ Thanks for your answer, but as @Tunk-Fey said, I didn't understand dilog function because I haven't learnt it yet. $\endgroup$ – Venus May 27 '14 at 1:26

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.