I will start my with an example, followed by questions based on this example, and then general questions that could apply to a general case.
Calculate the sum : $\sum_{n=1}^\infty \frac 1 {n^6}$
The solution uses the function $$f(x)=\begin{cases} x(\pi-x),&x\in[0,\pi]\\ x(\pi+x) , &x\in[-\pi,0]\end{cases}$$ $f$ is an odd function, so all of the even fourier coefficients are $0$, and all of the coefficients of the odd terms will be imaginary. Calculating the coefficients gives : $$\hat f_0=\frac 1 {2\pi}\int_{-\pi}^{\pi}f(t)dt=0$$ since $f$ is odd and the domain is symmetric around $0$, for $n\neq0$ we get : $$\hat f_n=\frac 1 {2\pi}\int_{-\pi}^{\pi}f(t)e^{-int}dt=\frac 1 {2\pi}\int_{-\pi}^{\pi}f(t)(\cos(nt)+i\sin(nt))dt=\\ =\frac 1 {2\pi}\left[\int_{-\pi}^{\pi}f(t)(\cos(nt))dt+\int_{-\pi}^{\pi}f(t)(i\sin(nt))dt\right]=\frac 1 {2\pi}\int_{-\pi}^{\pi}f(t)(i\sin(nt))dt$$
since $f(t)\cos(nt)$ is odd and the integral over a symmetric domain around $0$ is $0$.
Using integration by parts twice we get $$\hat f_n=\frac {2i((-1)^n-1)} {\pi n^3}$$ which means $$\hat f_n=\begin{cases}0 ,&\in\mathbb{N}_{odd} \\ -\frac {4i} {\pi n^3}&\text{otherwise}\end{cases}$$ and we get $$f(x) \sim \sum_{n=-\infty}^{\infty} e^{inx}\hat f_n$$ Using Parseval's identity we eventually get $$\|f\|^2=\sum_{n=-\infty}^{\infty}-\frac {4i} {\pi (2n+1)^3}$$ from here , we can evaluate the norm (it's a simple integral) and do several algebraic manipulations to the series and find the sum we wanted. This is a fairly straightforward and simple method if you have the right $f(x)$ for your sum.
My questions are about the construction of $f$:
$f$ is an odd function. In my understanding, $f$ was constructed this way just to simplfy the calculations ? Could we choose $f$ to be an even function and the calculations would be as easy? Could we choose $f$ to be neither even nor odd and still be able to calculate it simply? $f$ is a polynomial of degree 2. This means we will integrate by parts twice,resulting in a fourier coefficient in a degree of $\frac 1 {n^3}$. So in the case we want to calculate $\frac 1 {n^6}$ we will get the right degree from Parseval's identity. Does the fact that $f$ is even or odd affects this stage? For example, I've seen the even function $$f(x)=x^2$$ on the interval $[−π,π]$ gives coefficients which are of degree 2.
What if we wanted to calculate $$\sum _{n=1}^\infty \frac 1 {n^{2k}}$$ Would we choose $f(x)$ to be of degree $k$ and use the same method?
Would another way to calcultate $$\sum _{n=1}^\infty \frac 1 {n^{k}}$$ be to pick $f$ polynomial of degree $(k-1)$ that converges pointwise at $0$ (or maybe other "nice" number like $0.5\pi,\pi$)? Is it even possible? The idea is to evaluate $$f(0)=\sum _{n=-\infty}^{\infty} \hat f_n e^{in0}$$
The next question is related to Paseval's Identity:
- Can Pasevals Identity be generalized to $L_p$ norms? $$||f||_{L_p}^p=\sum_{n=-\infty}^{\infty}|\hat f_n|^p$$
If so, if the fourier series converges uniformly$^*$ to $f$, can we calculate sums using this generalized identity?
$(*)$-I haven`t gotten deep into Functional Analysis, but it does make sense to me that if the convergence is uniform then there is convergence to the function in any $L_p$ norm
I hope my questions are clear and I`m sorry for the long question.
Edit: Note that I`m not interested in help with the specific example I wrote down, I understand how to calculate the sum. I'm more interested in the general case and the construction of $f$, and the properties of $f$ that allows us to calculate the sum easily. The specific series I adressed to is only an example to show the method I was using. (And hopefully to be more clear with my question).Any answers on this specific case that can be generalized would be appriciated as well.