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?


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. $\endgroup$ – Alfred Yerger Dec 2 '15 at 20:36
  • $\begingroup$ @AlfredYerger Thanks! And pleased to hear that this becomes part of your tool kit!! $\endgroup$ – Mark Viola Dec 2 '15 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 '15 at 21:52
  • $\begingroup$ Nice work. Thanks. +1 $\endgroup$ – Kay K. Dec 2 '15 at 22:18
  • $\begingroup$ @KayK. Thanks.. $\endgroup$ – E.H.E Dec 2 '15 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. $\endgroup$ – Gregory Grant Dec 2 '15 at 21:37
  • $\begingroup$ Nicely done +1. $\endgroup$ – Kay K. Dec 2 '15 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 '15 at 10:27
  • $\begingroup$ @PatrickT He said he's looking for tricks, that is a trick. $\endgroup$ – Gregory Grant Dec 3 '15 at 12:23
  • $\begingroup$ Yes, he said "smart tricks" which this certainly is... $\endgroup$ – PatrickT Dec 3 '15 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 '15 at 20:48
  • $\begingroup$ Back at you. +1 $\endgroup$ – Mark Viola Dec 2 '15 at 22:19

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.