TITLE: Exact Sum of Series
QUESTION [24 upvotes]: 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?

REPLY [22 votes]: 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 $x=\frac{1}{\sqrt{3}}$
$$\sum_{n=1}^{\infty }\frac{(-1)^n}{(2n+3)3^n}=\frac{8}{3}-\frac{\sqrt{3}\pi}{2}$$