How can I calculate this series? How can I calculate the series
$\displaystyle{%
\sum_{n=0}^{\infty }{\left(-1\right)^{n}
\over \left(2n + 1\right)\left(2n + 4\right)}\,\left(1 \over 3\right)^{n + 2}\
{\large ?}}$
 A: For $x\in (0,1)$ we have
\begin{eqnarray}
f(x)&=&\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)(2n+4)}x^{2n}=\frac13\sum_{n=0}^\infty\left(\frac{(-1)^n}{2n+1}x^{2n}-\frac{(-1)^n}{2n+4}x^{2n}\right)\\
&=&\frac{1}{3x}\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}x^{2n+1}-\frac{1}{3x^4}\sum_{n=0}^\infty\frac{(-1)^n}{2n+4}x^{2n+4}\\
&=&\frac{1}{3x}\int_0^x\sum_{n=0}^\infty(-1)^nt^{2n}\,dt-\frac{1}{3x^4}\int_0^xt^3\sum_{n=0}^\infty(-1)^nt^{2n}\\
&=&\frac{1}{3x}\int_0^x\frac{1}{1+t^2}\,dt-\frac{1}{3x^4}\int_0^x\frac{t^3}{1+t^2}\,dt\\
&=&\frac{1}{3x}\arctan x-\frac{1}{3x^4}\int_0^x\left(t-\frac{t}{1+t^2}\right)\,dt\\
&=&\frac{1}{3x}\arctan x-\frac{1}{3x^4}\left[\frac{x^2}{2}-\frac12\ln(1+x^2)\right].
\end{eqnarray}
Thus
\begin{eqnarray}
\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)(2n+4)}\left(\frac13\right)^{n+2}&=&\frac19f\left(\frac{1}{\sqrt{3}}\right)=\frac{1}{9\sqrt{3}}\arctan\frac{1}{\sqrt{3}}-\frac13\left[\frac16-\frac12\ln\frac43\right]\\
&=&\frac{\pi}{54\sqrt{3}}-\frac{1}{18}+\frac16\ln\frac43.
\end{eqnarray}
A: Hint: There are a total of $4$ sums to be calculated. So I'll show you how to calculate $1$ of them, since they are all similar. Let $a = \left(\frac13\right)^{0.5}$, then the $1^{st}$ sum is $$S = \sum_{k=0}^\infty\dfrac{a^{4k+1}}{4k+1} = \sum_{k=0}^\infty\int_0^ax^{4k}dx = \int_0^a\sum_{k=0}^\infty x^{4k}dx = \int_0^a\dfrac1{1 - x^4}dx$$ which can be done by partial fraction decomposition. So the other $3$ sums can be done similarly, as they are of the same type.
A: $\newcommand{\+}{^{\dagger}}%
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 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
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$\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}
