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How to compute the following series:

$$\sum_{n=1}^{\infty}\frac{n+1}{2^nn^2}$$

I tried

$$\frac{n+1}{2^nn^2}=\frac{1}{2^nn}+\frac{1}{2^nn^2}$$

The idea is using Riemann zeta function

$$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$$

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.

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    $\begingroup$ 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
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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}

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    $\begingroup$ Nice explanation. +1 $\endgroup$ – Random Variable May 27 '14 at 1:24
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    $\begingroup$ @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
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    $\begingroup$ @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
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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}

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  • $\begingroup$ 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
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    $\begingroup$ @Integrals I know that but I haven't found a convenience answer yet $\endgroup$ – Venus May 26 '14 at 23:16
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    $\begingroup$ 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
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    $\begingroup$ 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
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\sum_{n = 1}^{\infty}{n + 1 \over 2^{n}\,n^{2}}:\ {\large ?}}$

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

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

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  • $\begingroup$ 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

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