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.

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

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

