Sum of alternating series Looking to find the sum of the following series:
$$\sum^\infty_{n=1}\frac{(-1)^n}{(2n+1)3^n}$$
It converges due to the Alternating Series Test.
Prior attempts had me isolating the numerator and the $3^n$ in one fraction, and using a geometric series. However, this proved fruitless since it left the other part of the fraction with a denominator that implies divergence. I am looking to help a calculus student who turned to me for advice, but I myself have not been worked with converging series in almost a decade, and so I'm not as quick with the material as I used to be.
 A: Note that for $|t|\lt 1$ we have 
$$\frac{1}{1+t^2}=1-t^2+t^4-t^6+\cdots.$$
Integrating term by term from $0$ to $x$ we get
$$\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots.$$
Comparing with our given series, we see that in our given series the first term is missing. Note also that 
$$\frac{1}{3^n}=\sqrt{3}\frac{1}{(\sqrt 3)^{2n+1}}.$$
Now put things together. You will find that our series has sum a close relative of $\arctan(1/\sqrt{3})$. Recall this is $\frac{\pi}{6}$. 
A: To do this directly from geometric series--in case you don't immediately recall that ${1\over 1+x^2}$ has $\arctan(x)$ as an elementary anti-derivative--you can do
$$\sum_{n=0}^\infty x^{2n}={1\over 1-x^2},\quad |x|<1$$
This will eventually flip to arctan because of the change of variables $x\mapsto ix$, which flips ${1\over 1-x^2}\mapsto {1\over 1+x^2}$.
Then integrate both sides
$$\sum_{n=0}^\infty {1\over 2n+1}x^{2n+1}=\int_0^x{dt\over 1-t^2}$$
using partial fractions you get this is ${1\over 2}\log\left({1+x\over 1-x}\right)+C$, choosing the initial term to be $0$ as in the integral gives just ${1\over 2}\log\left({1+x\over 1-x}\right)$.
Then plugging in $x={i\over \sqrt{3}}$ which clearly has $|x|<1$ gives
$${i\over\sqrt{3}}\sum_{n=0}^\infty {(-1)^{n}\over (2n+1)3^{n}}$$
Reducing gives
$$\sum_{n=0}^\infty {(-1)^n\over (2n+1)3^n}={\sqrt{3}\over 2i}\log\left({1+{i\over \sqrt{3}}\over 1-{i\over \sqrt{3}}}\right)$$
Taking out the initial $n=0$ term gives

$$\sum_{n=1}^\infty {(-1)^n\over (2n+1)3^n}=\sqrt{3}\cdot\underbrace{{1\over 2i}\log\left({1+{i\over \sqrt{3}}\over 1-{i\over \sqrt{3}}}\right)}_{\text{this is an arctan}}-1$$

Finally we note that $\arctan x = {1\over 2i}\log\left({1+ix\over 1-ix}\right)$, which leaves us with $\sqrt{3}\arctan\left({1\over\sqrt 3}\right)-1={\pi\over 2\sqrt{3}}-1$
A: To find the sum consider the series

$$ \sum_{k=0}^{\infty} x^{2n} = \frac{1}{1-x^2} $$

Integrate both sides from zero to $x$ and then substitute $x=i/\sqrt{3}$ and simplify.
Notes: To evaluate the integral just use the change of variables $ t=\sin(u) $ which gives

$$ I = \int_{0}^{x} \frac{dt}{1-t^2} = \int_{0}^{\arcsin(x)}\sec(u)du=\dots\,. $$

You need the integral

$$ \int \sec(u)du = \ln(\sec(x)+\tan(x)). $$

