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- #1

\frac{\pi^2}{3} + 4\sum_{n = 1}^{\infty}\frac{(-1)^n}{n^2}\cos n\theta.

$$

Let's look at $f\left(\frac{\pi}{2}\right) = \frac{\pi^2}{4}$.

\begin{alignat*}{3}

\frac{\pi^2}{3} + \sum_{n = 1}^{\infty}\frac{(-1)^n}{n^2}\cos\frac{n\pi}{2} & = & \frac{\pi^2}{3} + \left(\frac{-1}{4} + \frac{1}{16} - \frac{1}{36} + \cdots\right)\\

\frac{1}{4}\sum_{n = 1}^{\infty}\frac{(-1)^n}{n^2} & = & \frac{\pi^2}{4} - \frac{\pi^2}{3}\\

& = &

\end{alignat*}

This isn't going to equate to

$$

\sum_{n = 1}^{\infty}\frac{(-1)^n}{n^2} = -\frac{\pi^2}{12}

$$

What is wrong?