As given, the expression for\(\pi = 2\sqrt 3 \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{(2n + 1){3^n}}}} \).
By using power series expansion \({\tan ^{ - 1}}(x) = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{{{x^{2n + 1}}}}{{2n + 1}}\)
Replace \(x\) for\(\frac{1}{{\sqrt 3 }}\), then the series expressed as follows
\({\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right) = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{{{{\left( {\frac{1}{{\sqrt 3 }}} \right)}^{2n + 1}}}}{{2n + 1}}\)
\({\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right) = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{1}{{{{(\sqrt 3 )}^{2n + 1}}2n + 1}}\)
\({\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right) = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{1}{{(\sqrt 3 ) \cdot {{(\sqrt 3 )}^{2n}} \cdot 2n + 1}}\)
\(\frac{\pi }{6} = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{1}{{(\sqrt 3 ) \cdot \left( {{3^n}} \right) \cdot 2n + 1}}\)
Further, multiply both side by 6
\(\begin{aligned}\pi &= \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{6}{{(\sqrt 3 ) \cdot \left( {{3^n}} \right) \cdot (2n + 1)}}\\\pi &= \frac{6}{{\sqrt 3 }}\sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{1}{{{3^n} \cdot (2n + 1)}}\\\pi &= \frac{6}{{\sqrt 3 }}\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{(2n + 1) \cdot {3^n}}}} \end{aligned}\)
Therefore, sum of the infinite series is\(\pi = \frac{6}{{\sqrt 3 }}\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{(2n + 1) \cdot {3^n}}}} \).
