The function provided is $f(x) = x$ for $0\leq x\leq 2$.
I need to find a Fourier sine series that represents f(x) by extending it outside this interval. Then I have to evaluate the series: $$\sum _{k=1}^{\infty }\frac {\left( -1\right) ^{k-1}} {k^2} $$
I obtained b(n) for odd extensions of $f(x) = x$ (Refer to picture attached $b(n)$) but I'm unsure of how to continue. How do I use this result of $b(n)$ to evaluate the series in question?
Let symbol $\{f(x)\}_a^b$ mean a function $f(x)$ defined on an interval $[a,b]$ and $(b-a)$-periodically extended outside this interval with the convention $$ \{f(a)\}_a^b=\{f(b)\}_a^b=\frac{f(a)+f(b)}2. $$
As you already know: $$ \{x\}_{-2}^{+2}=-\frac4\pi\sum_{n=1}^\infty\frac{(-1)^n}n\sin\frac{n\pi x}2. $$
Integration of both sides of the equation results in: $$ \left\{\frac{x^2}2\right\}_{-2}^{+2}-C=\frac8{\pi^2}\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\cos\frac{n\pi x}2.\tag1 $$
To determine the integration constant $C$ one integrates both sides of the equation from $-2$ to 2: $$ \left[\frac{x^3}6-Cx\right]_{-2}^{+2}=0\implies C=\frac23. $$
Substituting in (1) the value of $C$ and evaluating the expression at $x=0$ one obtains: $$ \sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}=\frac23\frac{\pi^2}{8}=\frac{\pi^2}{12}. $$
Note that the same result can be obtained much easier by using the Fourier transform of the even function $f(x)=\{x^2\}_{-2}^{+2}$.