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?
 A: 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 $\log(1+x)$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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\begin{align}
&\bbox[10px,#ffd]{\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over
\pars{2n + 1}\pars{2n + 2}3^{n + 1}}} =
{1 \over 3}\sum_{n = 0}^{\infty}{\pars{-1/3}^{n} \over 2n + 1} - {1 \over 3}\sum_{n = 0}^{\infty}
{\pars{-1/3}^{n} \over 2n + 2}
\\[5mm] = &\
{1 \over 3}\sum_{n = 0}^{\infty}{\pars{-1/3}^{n} \over 2n + 1} + \sum_{n = 1}^{\infty}{\pars{-1/3}^{n} \over 2n}
\\[5mm] = &\
{1 \over 3}\,{\root{3} \over \ic}\sum_{n = 0}^{\infty}{\pars{\ic/\root{3}}^{2n + 1} \over 2n + 1} +
\sum_{n = 1}^{\infty}{\pars{\ic/\root{3}}^{2n} \over 2n}
\\[5mm] = &\
{1 \over 3}\,{\root{3} \over \ic}\sum_{n = 1}^{\infty}{\pars{\ic/\root{3}}^{n} \over n}\,{1 - \pars{-1}^{n} \over 2} +
\sum_{n = 1}^{\infty}{\pars{\ic/\root{3}}^{n} \over n}
\,{1 + \pars{-1}^{n} \over 2}
\\[5mm] = &\
{\root{3} \over 3}\,\Im\sum_{n = 1}^{\infty}{\pars{\ic/\root{3}}^{n} \over n} +
\Re\sum_{n = 1}^{\infty}{\pars{\ic/\root{3}}^{n} \over n}
\end{align}

Since
  $\ds{\sum_{n = 1}^{\infty}{\pars{\ic/\root{3}}^{n} \over n} =
-\ln\pars{1 - {\root{3} \over 3}\,\ic} =
-\,{1 \over 2}\,\ln\pars{4 \over 3} + {\pi \over 6}\,\ic}$:

\begin{align}
&\bbox[10px,#ffd]{\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over
\pars{2n + 1}\pars{2n + 2}3^{n + 1}}} =
{\root{3} \over 3}\,{\pi \over 6} -
{1 \over 2}\ln\pars{4 \over 3}
\\[5mm] = &\
\bbx{{\root{3} \over 18}\,\pi - {1 \over 2}\ln\pars{4 \over 3}}
\approx 0.1585
\end{align}
