Evaluate $1-\frac{1}{3\cdot 3}+\frac{1}{5\cdot 3^2}-\frac{1}{7\cdot 3^3}+\cdots$ My attempt
$$1-\frac{1}{3\cdot 3}+\frac{1}{5\cdot 3^2}-\frac{1}{7\cdot 3^3}+\cdots=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n-1)3^{n-1}}$$
By Leibniz alternative test for convergence. It is a convergent alternative series. How do I evaluate this limit?
 A: $$
\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n-1)3^{n-1}}=\sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)3^{n}}=\sqrt{3}\sum_{n=0}^\infty \frac{(-1)^{n}(1/\sqrt{3})^{2n+1}}{(2n+1)}
$$
$$
=\sqrt{3}\arctan(1/\sqrt{3}) = \frac{\sqrt{3}\pi}{6}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{n = 1}^{\infty}{\pars{-1}^{n - 1} \over \pars{2n - 1}3^{n - 1}} = &
-3\sum_{n = 1}^{\infty}{\pars{-1/3}^{n} \over 2n - 1} =
-\,\root{3}\ic\sum_{n = 1}^{\infty}{\pars{\ic/\root{3}}^{2n -1} \over 2n - 1}
\\[5mm] & =
-\root{3}\ic\sum_{n = 1}^{\infty}{\pars{\ic/\root{3}}^{n} \over n}\,{1^{n} - \pars{-1}^{n} \over 2}
\\[5mm] & =
\root{3}\,\Im\sum_{n = 1}^{\infty}{\pars{\ic/\root{3}}^{n} \over n}
\\[5mm] & =
-\root{3}\Im\ln\pars{1 - {\root{3} \over 3}\,\ic}
\\[5mm] & =
-\root{3}\arctan\pars{-\,{\root{3} \over 3}} 
\\[5mm] & =
\bbx{{\root{3} \over 6}\,\pi}
\end{align}
