The first series is : $$\sum_{n=1}^{\infty }(-1)^n\frac{4}{(n\pi )^2}\{(\cos(A(n\pi) )-\cos(B(n\pi ))\left. \right \}=(A^2-B^2)$$ Where $A$, $B$ are positive real numbers less than $1$.
I need a proof of the previous identity.
The second series is: $$\sum_{n=1}^{\infty }\frac{4}{(n\pi )^2}\{(\cos(A(n\pi) )-\cos(B(n\pi ))\left. \right \}=$$ I need to know what the series equals.
It is sufficient to recognize some well-known Fourier series (the triangle wave and the sawtooth wave). Since: $$\frac{\cos(An\pi)-\cos(Bn\pi)}{n \pi}=\int_{A}^{B}\sin(n\pi x)\,dx,\tag{1} $$ we have that the first series equals: $$4\int_{A}^{B}\sum_{n=1}^{+\infty}(-1)^n\frac{\sin(n\pi x)}{n\pi}\,dx =4\int_{A}^{B}-\frac{x}{2}\,dx=A^2-B^2=(A-B)(A+B),\tag{2}$$ while the second series equals: $$4\int_{A}^{B}\sum_{n=1}^{+\infty}\frac{\sin(n\pi x)}{n\pi}\,dx =4\int_{A}^{B}\frac{1-x}{2}\,dx=(A-B)(A+B-2).\tag{3}$$