<s>[INST] Need to compute $\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+2)3^{n+1}}$. Is my solution correct? $$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+2)3^{n+1}}$$
Since $$\tan^{-1}x=\int \frac{1}{1+x^{2}} dx=\int (1-x^{2}+x^{4}+...)dx=x-\frac{x
^3}{3}+\frac{x^5}{5}+...$$
so$$\int \tan^{-1}x dx=\int (x-\frac{x^3}{3}+\frac{x^5}{5}+...)dx=\frac {x^2}{2}-\frac{x^4}{3\cdot4}+\frac {x^6}{5\cdot 6}+...=\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+2}}{(2n+1)(2n+2)}$$
Therefore, $$\int_{0}^{1/{\sqrt 3}}\tan^{-1}xdx=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)(2n+2)3^{n+1}}$$
$$\Rightarrow \frac{\tan^{-1}(\frac{1}{\sqrt 3})}{\sqrt 3}-\frac{\ln(1+\frac{1}{3})}{2}=\frac{\pi}{6\sqrt3}-\frac{\ln\frac{4}{3}}{2}$$
Is my method correct?
 [/INST] You series equals 
$$ \frac{1}{3}\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)3^n}-\sum_{n\geq 0}\frac{(-1)^n}{(2n+2)3^{n+1}} $$
or
$$ \frac{1}{\sqrt{3}}\arctan\frac{1}{\sqrt{3}}+\frac{1}{2}\sum_{n\geq 1}\frac{(-1)^n}{n 3^n}=\frac{\pi}{6\sqrt{3}}-\frac{1}{2}\log\left(1+\frac{1}{3}\right). $$
No integrals, just partial fraction decomposition, reindexing and the Maclaurin series of $\arctan(x)$ and $\log(1+x)$.
</s>