The following question is purely my curiosity. During my calculation to answer @Chris'ssis's question in chat room I encountered this series $$\sum_{n=0}^\infty\; \frac{\left(2-\sqrt{3}\right)^{2n+1}}{(2n+1)^2\quad\ \,\,}$$ and I knew how to find its closed-form. Now, I am interested in knowing the closed-form of the following series $$\sum_{n=0}^\infty\;(-1)^n \frac{\left(2-\sqrt{3}\right)^{2n+1}}{(2n+1)^2\quad\ \,\,}$$ Is it possible to find its closed-form? I have a hunch it has a closed-form since the first series has it. Honestly, I am totally clueless about this one. Any idea? Any help would be appreciated. Thanks in advance.

  • $\begingroup$ Would you consider an answer using the polylog function to be closed-form? $\endgroup$ Dec 5, 2014 at 16:04
  • $\begingroup$ @RobertIsrael Yes, you may post it if you know it (I believe so). ^^ $\endgroup$
    – Venus
    Dec 5, 2014 at 16:06
  • $\begingroup$ Mathematica says $$-\left(\sqrt{3}-2\right) \, _3F_2\left(\frac{1}{2},\frac{1}{2},1;\frac{3}{2},\frac{3}{2};-7+4 \sqrt{3}\right)$$ Yuckk!! $\endgroup$ Dec 5, 2014 at 17:47
  • $\begingroup$ @Integrator True! $\endgroup$
    – Venus
    Dec 5, 2014 at 17:50
  • $\begingroup$ @Venus: is there any reason not to use $$\sum_{n=0}^\infty\;(-1)^n \frac{\left(2-\sqrt{3}\right)^{2n+1}}{(2n+1)^2\quad\ \,}$$ so that there is no long bar at the left of the fraction? $\endgroup$
    – robjohn
    Dec 6, 2014 at 0:11

2 Answers 2


First Approach

Using PolyLog[2,x], aka $\mathrm{Li}_2(x)$, we can compute the sum $$ \begin{align} \sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)^2} &=\mathrm{Li}_2(x)-\tfrac14\mathrm{Li}_2(x^2) \end{align} $$ Therefore, we have $$ \begin{align} \sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)^2} &=-i\sum_{n=0}^\infty\frac{(ix)^{2n+1}}{(2n+1)^2}\\ &=-i\left[\mathrm{Li}_2(ix)-\tfrac14\mathrm{Li}_2(-x^2)\right] \end{align} $$ Plugging in $x=2-\sqrt3$ gives $$ \begin{align} \sum_{n=0}^\infty(-1)^n\frac{(2-\sqrt3)^{2n+1}}{(2n+1)^2} &=-i\left[\mathrm{Li}_2(i(2-\sqrt3))-\tfrac14\mathrm{Li}_2(4\sqrt3-7)\right]\\ &\doteq0.26586495827930698269 \end{align} $$ The last value is computed using Mathematica 8 from the $\mathrm{Li}_2$ formula:

-I(PolyLog[2,I(2 - Sqrt[3])]-1/4PolyLog[2,4Sqrt[3]-7])

Second Approach

Using the Lerch Transcendent, $$ \Phi(z,s,a)=\sum_{k=0}^\infty\frac{z^k}{(a+k)^s} $$ we get $$ \begin{align} \sum_{n=0}^\infty(-1)^n\frac{(2-\sqrt3)^{2n+1}}{(2n+1)^2} &=\frac{2-\sqrt3}4\Phi\left(4\sqrt3-7,2,\tfrac12\right)\\ &\doteq0.26586495827930698269 \end{align} $$ The last value is computed using Mathematica 8 from the $\Phi$ formula:


Third Approach

In my opinion, this gives the best closed form.

As in this answer, we can look at a related generating function. $$ \sum_{k=0}^\infty(-1)^k\frac{t^{2k+1}}{2k+1}=\arctan(t)\tag{1} $$ Dividing $(1)$ by $t$ and integrating gives $$ \begin{align} \hspace{-1.5cm} \sum_{k=0}^\infty(-1)^k\frac{(2-\sqrt3)^{2k+1}}{(2k+1)^2} &=\int_0^{2-\sqrt3}\frac{\arctan(t)}{t}\mathrm{d}t\tag{2a}\\ &=\int_0^{\pi/12}x\,\mathrm{d}\log(\tan(x))\tag{2b}\\ &=\frac\pi{12}\log(2-\sqrt3)-\int_0^{\pi/12}\log(\tan(x))\,\mathrm{d}x\tag{2c}\\ &=\frac\pi{12}\log(2-\sqrt3)+2\int_0^{\pi/12}\sum_{k=0}^\infty\frac{\cos((4k+2)x)}{2k+1}\,\mathrm{d}x\tag{2d}\\ &=\frac\pi{12}\log(2-\sqrt3)+\sum_{k=0}^\infty\frac{\sin((2k+1)\pi/6)}{(2k+1)^2}\tag{2e}\\ &=\frac\pi{12}\log(2-\sqrt3)+\frac12\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2}+\frac32\sum_{k=0}^\infty\frac{(-1)^k}{(6k+3)^2}\tag{2f}\\ &=\frac\pi{12}\log(2-\sqrt3)+\frac23\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2}\tag{2g}\\ &=\frac\pi{12}\log(2-\sqrt3)+\frac23\mathrm{G}\tag{2h}\\[12pt] &\doteq0.26586495827930698269 \end{align} $$ Explanation:
$\text{(2a)}$: divide $(1)$ by $t$ and integrate
$\text{(2b)}$: substitute $t=\tan(x)$
$\text{(2c)}$: integrate by parts
$\text{(2d)}$: $\log(\tan(x))=-2\sum\limits_{k=0}^\infty\frac{\cos((4k+2)x)}{2k+1}$
$\text{(2e)}$: integrate
$\text{(2f)}$: details below
$\text{(2g)}$: $\frac12+\frac32\cdot\frac19=\frac23$
$\text{(2h)}$: $\mathrm{G}=\sum\limits_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2}$ is Catalan's Constant
The last value is computed using Mathematica 8 from $\text{(2h)}$:


Details of $\mathbf{(2f)}$

$\begin{align} \sin((2k+1)\pi/6)&=\left(\frac12,\quad1\ \ ,\frac12,-\frac12,\,-1\,,-\frac12,\frac12,\quad1\ \ ,\frac12,\dots\right)\\ &=\left(\frac12,-\frac12,\frac12,-\frac12,\ \ \frac12,-\frac12,\frac12,-\frac12,\frac12,\dots\right)\\ &+\left(\ 0\ ,\quad\frac32,\ 0\ ,\quad\,0\ ,-\frac32,\quad\,0\ ,\ 0\ ,\quad\frac32,\ 0\ ,\dots\right)\\ 2k+1 &=(\ \ \,1\,,\quad\ 3\,,\ \,5\,,\quad\ 7\,,\quad\ 9\,,\quad11,13,\ \ \ 15\,,17\,,\dots) \end{align}$

Note that the $\pm\frac32$ corrections are at positions where $2k+1$ is $3$ times an odd number. That is, $$ \sum_{k=0}^\infty\frac{\sin((2k+1)\pi/6)}{(2k+1)^2} =\frac12\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2} +\frac32\sum_{k=0}^\infty\frac{(-1)^k}{(6k+3)^2} $$ which justifies $\text{(2f)}$.

  • 7
    $\begingroup$ would the downvoter care to comment? $\endgroup$
    – robjohn
    Dec 5, 2014 at 22:53
  • $\begingroup$ I don't understand this, but I won't down-vote for that! $\endgroup$ Dec 6, 2014 at 3:29
  • 1
    $\begingroup$ Very nice & detail answer robjohn. Just ignore the downvoter, he knows nothing the beauty of your answer. Don't let his action drive you mad. I'll choose your answer, I hope Chris's sis doesn't mind. :-) $\endgroup$
    – Venus
    Dec 6, 2014 at 3:30
  • $\begingroup$ @Integrator: what is it you don't understand? $\endgroup$
    – robjohn
    Dec 6, 2014 at 4:10
  • 1
    $\begingroup$ @robjohn (+1) Awesome ! :-) $\endgroup$
    – r9m
    Dec 26, 2014 at 7:17

The second serie has not elementary sum: see http://mathworld.wolfram.com/LerchTranscendent.html

  • 1
    $\begingroup$ It has an elementary sum if you allow Catalan's Constant. $\endgroup$
    – robjohn
    Dec 5, 2014 at 20:49

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