How can I calculate this series? How can I calculate the series
$\displaystyle{%
\sum_{n=0}^{\infty }{\left(-1\right)^{n}
\over \left(2n + 1\right)\left(2n + 4\right)}\,\left(1 \over 3\right)^{n + 2}\
{\large ?}}$

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$\ds{%
{\cal I} \equiv \sum_{n=0}^{\infty }{\left(-1\right)^{n}
\over \left(2n + 1\right)\left(2n + 4\right)}\,\left(1 \over 3\right)^{n + 2}:\ {\large ?}}$

Let's
$\ds{%
\fermi\pars{x} \equiv \sum_{n=0}^{\infty}
{\pars{-1}^{n}x^{n + 2}\over \pars{2n + 1}\pars{2n + 4}}}$ such that
$\ds{{\cal I} = \fermi\pars{1 \over 3}}$.
\begin{align}
\fermi'\pars{x}&=\half\sum_{n = 0}^{\infty}{\pars{-1}^{n}x^{n + 1}\over 2n + 1}
\\[3mm]
\fermi''\pars{x}&=\half\sum_{n = 0}^{\infty}
{\pars{-1}^{n}\pars{n + 1}x^{n}\over 2n + 1}
={1 \over 4}\sum_{n = 0}^{\infty}\pars{-1}^{n}x^{n}
+
{1 \over 4}\sum_{n = 0}^{\infty}{\pars{-1}^{n}x^{n}\over 2n + 1}
\\[3mm]&={1 \over 4}\,{1 \over 1 + x} + {1 \over 2x}\,\fermi'\pars{x}
\quad\imp\quad
\fermi''\pars{x} - {1 \over 2x}\,\fermi'\pars{x} = {1 \over 4}\,{1 \over 1 + x}
\\
{1 \over x^{1/2}}\,\fermi''\pars{x} - {\fermi'\pars{x} \over 2x^{3/2}}
&={1 \over 4}\,{1 \over x^{1/2}\pars{x + 1}}
\quad\imp\quad
\totald{\bracks{x^{-1/2}\fermi'\pars{x}}}{x} = {1 \over 4}\,{1 \over x^{1/2}\pars{x + 1}}
\end{align} 
$$
\fermi'\pars{x} = \half\,x^{1/2}\arctan\pars{x^{1/2}} + Cx^{1/2}
$$
where $C$ is a constant. However, $\fermi'\pars{x} \sim x/2$ when $x \sim 0$ which leads to $C = 0$.

Since $\fermi\pars{0} = 0$
\begin{align}
{\cal I} &= \fermi\pars{1 \over 3} = \half\int_{0}^{1/3}x^{1/2}\arctan\pars{x^{1/2}}\,\dd x
\\[3mm]&=
\left.{1 \over 3}\,x^{3/2}\arctan\pars{x^{1/2}}\right\vert_{0}^{1/3}
-
{1 \over 3}\int_{0}^{1/3}
x^{3/2}\,{1 \over \pars{x^{1/2}}^{2} + 1}\,\half\,x^{-1/2}\,\dd x
\\[3mm]&=
\underbrace{{1 \over 3}\,\pars{1 \over 3}^{3/2}\
\overbrace{\arctan\pars{\root{3} \over 3}}^{\ds{\pi/6}}}
_{\pi\root{\vphantom{\large A}3}/162}
-
{1 \over 6}\int_{0}^{1/3}\pars{1 - {1 \over x + 1}}\,\dd x
\\[3mm]&=
\color{#0000ff}{\large{\pi\root{3} \over 162} - {1 \over 18} + {1 \over 6}\ln\pars{4 \over 3}} \approx 0.0260
\end{align}
