Motivation: I have a question that asked me to find the Fourier series of some function $f(x) = \left\{\begin{array}A,\quad -1\lt x \leq 0 \\ Ax, \quad 0 \lt x \leq 1 \end{array}\right.$ periodic on $f(x+2)=f(x)$. They have given me what it is meant to equal in the end, so that we know if we have gone wrong.
Question: Is the below equality true? This is part of the Fourier series where my expression doesn't appear to simplify to their expression.
Expressions: Below is part of (Theirs = Mine) $$2\frac{A}{\pi^2}\sum \limits_{m=0}^\infty \frac{1}{(2m+1^2)}\cos\left[(2m+1)\pi x\right]=-\frac{A}{\pi^2}\sum \limits_{n=1}^\infty \left(\frac{(-1)^n}{n^2} - \frac{1}{n^2} \cos(n\pi x)\right)$$
Note:
That $(2m+1^2)$ is exactly how it is written down in their expression, so I am not sure if the $1^2$ is a typo e.g. $(2m+1)^2$ or not.
Many thanks
$$2\frac{A}{\pi^2}\sum \limits_{m=0}^\infty \frac{1}{(2m+1^2)}\cos\left[(2m+1)\pi x\right]\underset{\large\bf{=}}{?}-\frac{A}{\pi^2}\sum \limits_{n=1}^\infty \left(\frac{(-1)^n}{n^2} - \frac{1}{n^2} \cos(n\pi x)\right)$$
First note that the $(2m+1^2)$ should indeed be $(2m+1)^2$
Now we will note that $(-1)^n-1$ for $n=1,2,3,\dots$ is $-2,0,-2,0,\dots$ and we can take this $-2$ out the from the RHS, ensuring we are only counting for $n=1,3,5,\dots$ e.g. $n$ odd, giving:
$$\operatorname{RHS}=2\frac{A}{\pi^2}\sum \limits_{m=0}^\infty \left(\frac{1}{(2m+1)^2}\cos((2m+1)\pi x)\right)=\operatorname{LHS}$$