The Fourier series of the function,f(x) = 0, -π < x ≤ 0= π - x, 0 < x

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

\(\mathop \sum \limits_{n = 1}^\infty \frac{1}{{{n^2}}} = \frac{{{\pi ^2}}}{6}\)

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

\(\mathop \sum \limits_{n = 1}^\infty \frac{1}{{\left( {2n - 1} \right)^2}} = \frac{{{\pi ^2}}}{8}\)

The Fourier series of the function,f(x) = 0, -π < x ≤ 0= π - x, 0

1.	The Fourier series of the function,f(x) = 0, -π < x ≤ 0= π - x, 0 < x
A.	\(\mathop \sum \limits_{n = 1}^\infty \frac{1}{{{n^2}}} = \frac{{{\pi ^2}}}{6}\)
B.	\(\mathop \sum \limits_{n = 1}^\infty \frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{{n^2}}} = \frac{{{\pi ^2}}}{{12}}\)
C.	\(\mathop \sum \limits_{n = 1}^\infty \frac{1}{{\left( {2n - 1} \right)^2}} = \frac{{{\pi ^2}}}{8}\)
D.	\(\mathop \sum \limits_{n = 1}^\infty \frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{2n - 1}} = \frac{\pi }{4}\)
Answer» D. \(\mathop \sum \limits_{n = 1}^\infty \frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{2n - 1}} = \frac{\pi }{4}\)