Answer
$\frac{\pi\sqrt{3}}{6}-1$
Work Step by Step
In Example 7, we have $\tan ^{-1}x=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}$ where $|x|<1$.
Substituting $x=\frac{1}{\sqrt{3}}$,
$\tan^{-1}\frac{1}{\sqrt{3}}=\sum_{n=0}^\infty (-1)^n\frac{(1/\sqrt{3})^{2n+1}}{2n+1}$ (Simplify)
$\frac{\pi}{6}=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)3^n\sqrt{3}}$ (Multiply by $\sqrt{3}$)
$\frac{\pi\sqrt{3}}{6}=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)3^n}$
$\frac{\pi\sqrt{3}}{6}=1+\sum_{n=1}^\infty \frac{(-1)^n}{(2n+1)3^n}$
Thus,
$\sum_{n=1}^\infty \frac{(-1)^n}{(2n+1)3^n}=\frac{\pi\sqrt{3}}{6}-1$
