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The solution using the Fourier series of x and Parseval's Identity to prove that

[tex] \sum_{n=1}^{∞} \frac{1} {n^2} = \frac{\pi^2} {6} [/tex]

doesn't seem to work for exponents higher than 2, why?

For example, I found the Fourier coefficients of [itex] - \frac{x^2} {4} [/itex] to be

[itex] \frac{(-1)^{n+1}} {n^2} [/itex].

So if I use the same reasoning as in the Basel problem:

[tex] \sum_{n=1}^{∞} \frac{1} {n^4} = \frac{1} {\pi} \int_{-\pi}^{\pi} \frac{x^4} {16} dx = \frac{\pi^4} {40}[/tex]

which is not even close to the correct value.

How come when Euler does it, it works like magic, but when I try to do the same thing, it fails? Am I doing something wrong? Is Parseval's Identity valid only for the function f(x) = x?

[tex] \sum_{n=1}^{∞} \frac{1} {n^2} = \frac{\pi^2} {6} [/tex]

doesn't seem to work for exponents higher than 2, why?

For example, I found the Fourier coefficients of [itex] - \frac{x^2} {4} [/itex] to be

[itex] \frac{(-1)^{n+1}} {n^2} [/itex].

So if I use the same reasoning as in the Basel problem:

[tex] \sum_{n=1}^{∞} \frac{1} {n^4} = \frac{1} {\pi} \int_{-\pi}^{\pi} \frac{x^4} {16} dx = \frac{\pi^4} {40}[/tex]

which is not even close to the correct value.

How come when Euler does it, it works like magic, but when I try to do the same thing, it fails? Am I doing something wrong? Is Parseval's Identity valid only for the function f(x) = x?

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