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Prove some formulas for given infinite sums

Using the formula

    \[ \sum_{n=1}^{\infty} \frac{\cos (nx)}{n^2} = \frac{x^2}{4} - \frac{\pi x}{2} + \frac{\pi^2}{6} \qquad 0 \leq x \leq 2 \pi \]

prove that:

  1. \displaystyle{\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}}.
  2. \displaystyle{\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(2n-1)^3} = \frac{\pi^3}{32}}.

  1. Proof. Taking x = 0 in the given formula we have

        \begin{align*}  && \sum_{n=1}^{\infty} \frac{\cos (nx)}{n^2} &= \frac{x^2}{4} - \frac{\pi x}{2} + \frac{\pi^2}{6} \\[9pt]  \implies && \sum_{n=1}^{\infty} \frac{\cos (0)}{n^2} &= \frac{\pi^2}{6} \\[9pt]  \implies && \sum_{n=1}^{\infty} \frac{1}{n^2} &= \frac{\pi^2}{6}. \qquad \blacksquare \end{align*}

  2. Proof. Starting with the given equation we have for 0 \leq x \leq 2 \pi,

        \begin{align*}  && \sum_{n=1}^{\infty} \frac{\cos (nx)}{n^2} &= \frac{x^2}{4} - \frac{\pi x}{2} + \frac{\pi^2}{6} \\[9pt]  \implies && \int \sum_{n=1}^{\infty} \frac{\cos (nx)}{n^2} \, dx &= \int \left( \frac{x^2}{4} - \frac{\pi x}{4} + \frac{\pi^2}{6} \right) \, dx \\[9pt]  \implies && \sum_{n=1}^{\infty} \left(\int \frac{ \cos (nx)}{n^2} \, dx \right)&= \frac{x^3}{12} - \frac{\pi x^2}{4} + \frac{\pi^2 x}{6} \\[9pt]  \implies && \sum_{n=1}^{\infty} \frac{\sin (nx)}{n^3} &= \frac{x^3}{12} - \frac{\pi x^2}{4} + \frac{\pi^2 x}{6}. \end{align*}

    Take x = \frac{\pi}{2}, then we have

        \begin{align*}  && \sum_{n=1}^{\infty} \frac{\sin \left( \frac{\pi}{2} \cdot n \right)}{n^3} &= \frac{ \left(\frac{\pi}{2} \right)^3}{12} - \frac{ \pi \left( \frac{\pi}{2} \right)^2}{4} + \frac{\pi^2 \left( \frac{\pi}{2} \right)}{6} \\[9pt]  \implies && \sum_{n=1}^{\infty} \frac{\sin \left( n \pi - \frac{\pi}{2} \right)}{(2n-1)^3} &= \frac{\pi^3}{32} &(\sin (n \pi) = 0 \text{ for even }n)\\[9pt]  \implies && \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(2n-1)^3} &= \frac{\pi^3}{32}. \qquad \blacksquare \end{align*}

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