How can i evaluate this power series? $\sum_{n=0}^{\infty }\frac{1}{2n+1} \left (\frac{1}{3}  \right )^{n}\left ( -1 \right )^{n} $
it's solved by power series of arctan. is it possible the answer written by real number?
 A: Form this, for $|x|<1$  $$\ln(1+x)=x-\frac{x^2}2+\frac{x^3}3-\cdots\text{ and }\ln(1-x)=-x-\frac{x^2}2-\frac{x^3}3-\cdots$$
$$\implies\ln(1+x)-\ln(1-x)=?$$
Now observe that, $$\left(-\frac1{\sqrt3i}\right)^{2n+1}=-\left(\frac1{\sqrt3i}\right)^{2n+1}=-\frac1{\sqrt3i}\frac1{(\sqrt3i)^{2n}}=-\frac1{\sqrt3i}\frac1{3^n(-1)^n}=\frac i{\sqrt3}\frac{(-1)^n}{3^n}$$
So, the sum reduces to $$\frac i{\sqrt3}\ln\frac{1-\frac1{\sqrt3i}}{1+\frac1{\sqrt3i}}=\frac i{\sqrt3}\ln\frac{\sqrt3+i}{\sqrt3-i}=\frac i{\sqrt3}\ln\left(\frac{\sqrt3}2+i\frac12\right)$$
$$=\frac i{\sqrt3}\ln e^{(2n\pi i+ i\frac\pi3)}=\frac{i(2n\pi i+ i\frac\pi3)}{\sqrt3}=\frac{-2n\pi-\frac\pi3}{\sqrt3}$$
As we are interested in the principal value, $n=0$
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
&\color{#0000ff}{\large%
\sum_{n = 0}^{\infty}{1 \over 2n+1}\pars{1 \over 3}^{n}\pars{-1}^{n}}
=\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over 2n+1}\pars{1 \over \root{3}}^{2n}
=\root{3}\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over 2n+1}
\pars{1 \over \root{3}}^{2n + 1}
\\[3mm]&=\root{3}\int_{0}^{1/\root{3}}\sum_{n = 0}^{\infty}\pars{-1}^{n}x^{2n}\,\dd x
=\root{3}\int_{0}^{1/\root{3}}{\dd x \over 1 + x^{2}} = \root{3}\arctan\pars{1 \over \root{3}}
=\root{3}\,{\pi \over 6}
\\[3mm]&=\color{#0000ff}{\large{\root{3} \over 6}\,\pi}
\end{align}
