I have done some exercises on the Parseval's identity and I think it's quite straight forward. However, I came across this exercise and it made me confused. I'll explain:
The function is $f(x) = \pi x – x^2$ on the interval [0, 𝜋].
I shall find the sine series of the function and prove that: $$\sum_{n=1}^\infty \frac{1}{n^6} = \frac{\pi^6}{945}$$
Now, I found the correct Fourier coefficients and I get:
$$f(x) = \sum_{n=1}^\infty \frac{4}{\pi n^3}(1-(-1)^n)$$
Perfect. Now, using the Parseval's I want to calculate $|b_n|^2$ and here is where the problem occure. The problem is the $(1-(-1)^n)^2$ term. I tried to evaluate it:
- $(\frac{4}{\pi n^3}(1-(-1)^n)^2 = \frac{16}{\pi^2 n^6}(1-(-1)^n)^2 = \frac{16}{\pi^2 n^6}(1-2(-1)^n + (-1)^{2n}) = \frac{16}{\pi^2 n^6}(2-2(-1)^n)$ which is 0 for even n!
So what I get is: $$
|b_n|^2 =
\begin{cases}
\frac{64}{\pi^2(2k+1)^6}, & n \text{ odd} \\\\
0, & n \text{ even.}
\end{cases}
$$
Now, how does that help me prove what I should prove???
Best Answer
Since as you wrote
$$1-(-1)^n=\begin{cases}0\,,\,\,n\;\text{is even}\\{}\\ 2\;,\;\;n\;\text{is odd}\end{cases}$$
so that your sine series is
$$f(x)=\sum_{n=1}^\infty\frac{4\cdot2}{\pi(2n-1)^3}\sin nx=\frac8\pi\sum_{n=1}^\infty\frac1{(2n-1)^3}\sin nx$$
And now Parsival:
$$\frac1{2\pi}\int_0^\pi(\pi x-x^2)^2 dx=\frac{64}{\pi^2}\sum_{n=1}^\infty\frac1{(2n-1)^6}\implies \sum_{n=1}^\infty\frac1{(2n-1)^6}=\frac{\pi^6}{15\cdot64}$$
And finally:
$$\sum_{n=1}^\infty\frac1{n^6}=\sum_{n=1}^\infty\frac1{(2n-1)^6}+\sum_{n=1}^\infty\frac1{(2n)^6}=\frac{\pi^6}{15\cdot 64}+\frac1{64}\sum_{n=1}^\infty\frac1{n^6}\implies$$
$$\frac{63}{64}\sum_{n=1}^\infty\frac1{n^6}=\frac{\pi^6}{15\cdot 64}\implies\sum_{n=1}^\infty\frac1{n^6}=\frac{\pi^6}{945}$$
Check the above so that everything's clear...and correct.