I am a tutor at university, and one of my students brought me this question, which I was unable to work out. It is from a past final exam in calculus II, so any response should be very basic in what machinery it uses, although it may be complicated. The series is: $$\sum \limits_{n=1}^{\infty} \frac{(-1)^n}{(2n+3)(3^n)}.$$

Normally I'm pretty good with infinite series. It is clear enough to me that this sum converges. None of the kind of obvious rearrangements yielded anything, and I couldn't come up with any smart tricks in the time we had. I put it into Wolfram and got a very striking answer indeed. Wolfram reports the value to be $\frac{1}{6}(16-3\sqrt{3} \pi)$. It does this using something it calls the "Lerch Transcendent" (link here about Lerch). After looking around, I think maybe I can understand how the summing is done, if you knew about this guy and special values it takes.

But how could I do it as a calculus II student, never having seen anything like this monstrosity before?


4 Answers 4


Note that we can write the sum as $f(1/3)$ where

$$f(x)=x^{-3/2}\sum_{n=1}^{\infty}(-1)^n\frac{\left(\sqrt{x}\right)^{2n+3}}{2n+3}\tag 1$$

Now, denote the series in $(1)$ by $g(x)$. Then, we have

$$g'(x)=\frac12\sqrt{x}\,\,\sum_{n=1}^\infty (-1)^n x^n=- \frac{x^{3/2}}{2(x+1)} \tag 2$$

Integrating both sides of $(2)$ yields

$$g(x)=-\frac13 \sqrt{x}(x-3)-\arctan(\sqrt{x}) \tag 3$$

Now, simply substitute $g(x)$ in $(3)$ into $(1)$ and evaluate at $x=1/3$. Proceeding, we find


  • $\begingroup$ Ah this is a good idea. I didn't think to look for Taylor series we could use. This is a really good trick in general for evaluating infinite series then. I will remember this. $\endgroup$ Dec 2, 2015 at 20:36
  • $\begingroup$ @AlfredYerger Thanks! And pleased to hear that this becomes part of your tool kit!! $\endgroup$
    – Mark Viola
    Dec 2, 2015 at 20:38

depending on the geometric series $$\frac{1}{1+x}=\sum_{n=0}^{\infty }(-1)^nx^n$$ $$ \frac{1}{1+x}-1=\sum_{n=1}^{\infty }(-1)^nx^n$$ $$ \frac{-x}{1+x}=\sum_{n=1}^{\infty }(-1)^nx^n$$

$x\rightarrow x^2$ $$ \frac{-x^2}{1+x^2}=\sum_{n=1}^{\infty }(-1)^nx^{2n}$$

multiply by $x^2$ $$ \frac{-x^4}{1+x^2}=\sum_{n=1}^{\infty }(-1)^nx^{2n+2}$$ $$ \int_{0}^{x}\frac{-x^4}{1+x^2}dx=\int_{0}^{x}\sum_{n=1}^{\infty }(-1)^nx^{2n+2}dx$$ $$x-x^3/3-\tan^{-1}(x)=\sum_{n=1}^{\infty }\frac{(-1)^nx^{2n+3}}{2n+3}$$ divide by $x^3$ $$1/x^2-1/3-\tan^{-1}(x)/x^3=\sum_{n=1}^{\infty }\frac{(-1)^nx^{2n}}{2n+3}$$

now let $x=\frac{1}{\sqrt{3}}$ $$\sum_{n=1}^{\infty }\frac{(-1)^n}{(2n+3)3^n}=\frac{8}{3}-\frac{\sqrt{3}\pi}{2}$$

  • 1
    $\begingroup$ There you go. +1 $\endgroup$
    – Mark Viola
    Dec 2, 2015 at 21:52
  • $\begingroup$ Nice work. Thanks. +1 $\endgroup$
    – Kay K.
    Dec 2, 2015 at 22:18
  • $\begingroup$ @KayK. Thanks.. $\endgroup$
    – E.H.E
    Dec 2, 2015 at 22:20

I believe you can get there from $\tan^{-1}(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{2n+1}$.

Substitute $n=m+1$ to get


Then substitute $x=\frac{1}{\sqrt{3}}$.

Should yield $-3\sqrt{3}(\tan^{-1}\frac{1}{\sqrt3}-\frac13)$.

Or something like that.

  • $\begingroup$ @E.H.E Thank you, not sure I got the exact right answer but the method should work. $\endgroup$ Dec 2, 2015 at 21:37
  • $\begingroup$ Nicely done +1. $\endgroup$
    – Kay K.
    Dec 2, 2015 at 22:19
  • $\begingroup$ nice, but isn't the first line of your argument already a big part of the answer? $\endgroup$
    – PatrickT
    Dec 3, 2015 at 10:27
  • $\begingroup$ @PatrickT He said he's looking for tricks, that is a trick. $\endgroup$ Dec 3, 2015 at 12:23
  • $\begingroup$ Yes, he said "smart tricks" which this certainly is... $\endgroup$
    – PatrickT
    Dec 3, 2015 at 14:08

\begin{align} &f(x)=\sum_{k=1}^{\infty}x^{2n+2}=x^4\sum_{k=0}^{\infty}x^{2n}=\frac{x^4}{1-x^2}\\ &F(x)=\int f(x) dx=\frac{-x(x^2+3)}3+\frac{\ln {\frac{1+x}{1-x}}}2=\sum_{k=1}^{\infty}\frac{x^{2n+3}}{2n+3}\\ &G(x)=\frac{F(x)}{x^3}=\sum_{k=1}^{\infty}\frac{x^{2n}}{2n+3}=\frac{-(x^2+3)}{3x^2}+\frac{\frac1x\ln {\frac{1+x}{1-x}}}{2x^2}\\ &G\left(\frac{i}{\sqrt3}\right)=\sum_{k=1}^{\infty}\frac{\left(-\frac13\right)^n}{2n+3}=\frac{-(-\frac13+3)}{-1}+\frac{\frac{\sqrt3}{i}\ln {\left(\frac{\sqrt3+i}{2}\right)^2}}{2\cdot\left(-\frac13\right)}\\ &=\frac83-\frac{3\sqrt3}{i} \ln\left(e^{\frac{\pi i}6}\right)=\frac83-\frac{\sqrt3\pi}2 \end{align}

  • $\begingroup$ Now I realized that I did it a little weird (it would have been easier if I had kept $(-1)^n$), but I'll leave it as is just to show that this is also possible. $\endgroup$
    – Kay K.
    Dec 2, 2015 at 20:48
  • $\begingroup$ Back at you. +1 $\endgroup$
    – Mark Viola
    Dec 2, 2015 at 22:19

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