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The Fourier series of the function, $\begin{aligned}
f(x) &=0, &-& \pi &=\pi-x, & 0 \end{aligned}$ in the interval $[-\pi, \pi]$ is$ f(x)=\frac{\pi}{4}+\frac{2}{\pi}\left[\frac{\cos x}{1^{2}}+\frac{\cos 3 x}{3^{2}}+\ldots \ldots \ldots\right]+\left[\frac{\sin x}{1}+\frac{\sin 2 x}{2}+\frac{\sin 3 x}{3}+\ldots \ldots\right] $ The convergence of the above Fourier series at $x=0$ gives
(A) $\sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}$
$\text { (B) } \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{2}}=\frac{\pi^{2}}{12}$
(C) $\sum_{n=1}^{\infty} \frac{1}{(2 n-1)^{2}}=\frac{\pi^{2}}{8}$
$\text { (D) } \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2 n-1}=\frac{\pi}{4}$


Question ID - 156498 | Toppr Answer

The Fourier series of the function, $\begin{aligned}
f(x) &=0, &-& \pi &=\pi-x, & 0 \end{aligned}$ in the interval $[-\pi, \pi]$ is$ f(x)=\frac{\pi}{4}+\frac{2}{\pi}\left[\frac{\cos x}{1^{2}}+\frac{\cos 3 x}{3^{2}}+\ldots \ldots \ldots\right]+\left[\frac{\sin x}{1}+\frac{\sin 2 x}{2}+\frac{\sin 3 x}{3}+\ldots \ldots\right] $ The convergence of the above Fourier series at $x=0$ gives
(A) $\sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}$
$\text { (B) } \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{2}}=\frac{\pi^{2}}{12}$
(C) $\sum_{n=1}^{\infty} \frac{1}{(2 n-1)^{2}}=\frac{\pi^{2}}{8}$
$\text { (D) } \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2 n-1}=\frac{\pi}{4}$

1 Answer - 5876 Votes

3537

Answer Key : (C) -

$\sum_{n=1}^{\infty} \frac{1}{(2 n-1)^{2}}=\frac{\pi^{2}}{8}$



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