Exact Sum of Series 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?
 A: \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}
A: 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
$$f(1/3)=\frac83-\frac{\pi\sqrt{3}}{2}$$
A: 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 $x=\frac{1}{\sqrt{3}}$
$$\sum_{n=1}^{\infty }\frac{(-1)^n}{(2n+3)3^n}=\frac{8}{3}-\frac{\sqrt{3}\pi}{2}$$
A: 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
$$\sum_{m=-1}^{\infty}\frac{(-1)^{m+1}}{2m+3}x^{2m+3}.$$
Then substitute $x=\frac{1}{\sqrt{3}}$.
Should yield $-3\sqrt{3}(\tan^{-1}\frac{1}{\sqrt3}-\frac13)$.
Or something like that.
