Answer
$\dfrac{\pi}{6}$
Work Step by Step
Since, we know that the Maclaurin Series for $ \tan^{-1} x$ is defined as:
$ \tan^{-1} x=\Sigma_{n=0}^\infty (-1)^n\dfrac{x^{2n-1}}{(2n-1)}=x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-\dfrac{x^7}{7}+...$
Thus, we have the form of the given series as $ \tan^{-1} x$ series.
Here, in the given problem $x=\dfrac{1}{\sqrt 3}$
Thus, the sum of the series is:
$ \tan^{-1} (\dfrac{1}{\sqrt 3})=\tan^{-1} (\tan (\dfrac{\pi}{6}))=\dfrac{\pi}{6}$
