But the denominator is so large

We know that $\sum\limits_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1} = \arctan(x) , \forall x\in]-1,1[$ . Therefore, $\sum\limits_{n=0}^\infty (-1)^n \frac{x^{2n+2}}{2n+1} = x\arctan(x) , \forall x\in]-1,1[$ . It is clear that $\sum\limits_{n=0}^\infty (-1)^n \frac{1}{3^{n+1}(2n+1)}=\sum\limits_{n=0}^\infty (-1)^n \frac{\frac{1}{3^{n+1}}}{(2n+1)}=\sum\limits_{n=0}^\infty (-1)^n \frac{{\left( \frac{1}{3} \right)}^{n+1}}{(2n+1)}=$ $\sum\limits_{n=0}^\infty (-1)^n \frac{{\left( \frac{1}{\sqrt{3}} \right)}^{2(n+1)}}{(2n+1)}=\sum\limits_{n=0}^\infty (-1)^n \frac{{\left( \frac{1}{\sqrt{3}} \right)}^{2n+2}}{(2n+1)}$ and, because $\frac{1}{\sqrt{3}} \in ]-1,1[$ , $\sum\limits_{n=0}^\infty (-1)^n \frac{{\left( \frac{1}{\sqrt{3}} \right)}^{2n+2}}{(2n+1)} =\frac{1}{\sqrt{3}}\arctan\left(\frac{1}{\sqrt{3}}\right)$ . So, we have that $\sum\limits_{n=0}^\infty (-1)^n \frac{1}{3^{n+1}(2n+1)}=\frac{1}{\sqrt{3}}\arctan\left(\frac{1}{\sqrt{3}}\right)=\frac{π\sqrt{3}}{18}$ .

Relevant wiki: Maclaurin Series

Consider the Maclaurin series for $\tan^{-1} x$ :

$\begin{aligned} \sum_{n=0}^\infty \frac {(-1)^n x^{2n+1}}{2n+1} & = \tan^{-1} x & \small \color{#3D99F6} \text{Multiplying both sides by }x \\ \sum_{n=0}^\infty \frac {(-1)^n\left(x^2\right)^{n+1}}{2n+1} & = x \tan^{-1} x & \small \color{#3D99F6} \text{Putting }x= \frac 1{\sqrt 3} \\ \sum_{n=0}^\infty \frac {(-1)^n}{3^{n+1}(2n+1)} & = \frac 1{\sqrt 3} \tan^{-1} \frac 1{\sqrt 3} \\ & = \frac 1{\sqrt 3} \times \frac \pi 6 \\ & = \boxed{\dfrac {\pi \sqrt 3}{18}} \end{aligned}$