The Fourier series of the function,
\[\begin{array}{*{20}{c}} {{\text{f}}\left( {\text{x}} \right) = 0,}&{ - \pi < {\text{x}} \leqslant 0} \\ {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \pi - {\text{x,}}}&{0 < {\text{x}} < \pi } \end{array}\]      in the interval $$\left[ { - \pi ,\,\pi } \right]$$  is $${\text{f}}\left( {\text{x}} \right) = \frac{\pi }{4} + \frac{2}{\pi }\left[ {\frac{{\cos {\text{x}}}}{{{1^2}}} + \frac{{\cos {\text{3x}}}}{{{3^3}}} + \,...} \right] + \left[ {\frac{{\sin {\text{x}}}}{1} + \frac{{\sin {\text{2x}}}}{2} + \frac{{\sin {\text{3x}}}}{3} + \,...} \right]$$
The convergence of the above Fourier series at x = 0 gives

A. $$\sum\limits_{{\text{n}} = 1}^\infty {\frac{1}{{{{\text{n}}^2}}} = \frac{{{\pi ^2}}}{6}} $$

B. $$\sum\limits_{{\text{n}} = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{{\text{n}} + 1}}}}{{{{\text{n}}^2}}} = \frac{{{\pi ^2}}}{{12}}} $$

C. $$\sum\limits_{{\text{n}} = 1}^\infty {\frac{1}{{{{\left( {{\text{2n}} - 1} \right)}^2}}} = \frac{{{\pi ^2}}}{8}} $$

D. $$\sum\limits_{{\text{n}} = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{{\text{n}} + 1}}}}{{{\text{2n}} - 1}} = \frac{\pi }{4}} $$

Answer: Option C