# 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) $\sum_{n=1}^{\infty} \frac{1}{(2 n-1)^{2}}=\frac{\pi^{2}}{8}$ $\text { (D) } \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2 n-1}=\frac{\pi}{4}$

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