MHB Investigating the Inequality of $\frac{\pi^2}{4}$ and $\frac{-\pi^2}{12}$

[Dustinsfl]

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

 

[Opalg]

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

It would have been a great deal easier to understand this question if you had thought to mention that this is the Fourier series of the function $f(\theta)= \theta^2.$

You have given the Fourier series correctly, and the function is equal to the sum of the series, so you should be starting with the equation $$\theta^2 = \frac{\pi^2}{3} + 4\sum_{n = 1}^{\infty}\frac{(-1)^n}{n^2}\cos n\theta.$$ When you put $\theta = \pi/2$ this becomes $$\frac{\pi^2}4 = \frac{\pi^2}{3} + 4\sum_{n = 1}^{\infty}\frac{(-1)^n}{n^2}\cos \bigl(\tfrac{n\pi}2\bigr).$$ The value of $\cos \bigl(\tfrac{n\pi}2\bigr)$ is zero when $n$ is odd, and it is $(-1)^k$ when $n=2k$ is even. Therefore $$\frac{\pi^2}4 = \frac{\pi^2}{3} + 4\sum_{k = 1}^{\infty}\frac{(-1)^k}{(2k)^2}.$$ Notice the $(2k)^2$ in the denominator!