Get the solution to "Are there any nice ways to prove thatζ(4)=∑n=1∞1n4=π490"

Cian Orr

Answered question

2022-02-27

Are there any nice ways to prove that

ζ (4) = \sum_{n = 1}^{\infty} \frac{1}{n^{4}} = \frac{π^{4}}{90}

jaewonlee0217fyv

Beginner2022-02-28Added 5 answers

If we substitute

π

for x in the Fourier trigonometric series expansion of

f (x) = x^{4}

, with

- π \leq x \leq π

x^{4} = \frac{1}{5} π^{4} + \sum_{n = 1}^{\infty} \frac{8 n^{2} π^{2} - 48}{n^{4}} \cos n π \cdot \cos n x

we obtain

π^{4} = \frac{1}{5} π^{4} + \sum_{n = 1} \frac{8 n^{2} π^{2} - 48}{n^{4}} \cos^{2} n π

= \frac{1}{5} π^{4} + 8 π^{2} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} - 48 \sum_{n = 1}^{\infty} \frac{1}{n^{4}}

Hence

\sum_{n = 1}^{\infty} \frac{1}{n^{4}} = \frac{π^{4}}{48} (- 1 + \frac{1}{5} + \frac{8}{6})

= \frac{π^{4}}{48} \cdot \frac{8}{15}

= \frac{1}{90} π^{4}

Halle Hansen

Beginner2022-03-01Added 5 answers

So, we know that

\sum_{n = - \infty}^{\infty} \frac{1}{{(z + n)}^{2}} = \frac{π^{2}}{\sin^{2} (π z)}

Differentiating this twice, we have

\sum_{n = - \infty}^{\infty} \frac{1}{{(z + n)}^{4}} = \frac{π^{4} (2 + \cos (2 π z))}{3 \sin^{4} (π z)}

\sum_{n = 1}^{\infty} (\frac{1}{{(z - n)}^{4}} + \frac{1}{{(z + n)}^{4}}) = \frac{π^{4} (2 + \cos (2 π z))}{3 \sin^{4} (π z)} - \frac{1}{z^{4}}

Note that the LHS of the above is analytic

z = 0

and hence

\sum_{n = 1}^{\infty} \frac{1}{n^{4}} = lim_{z \to 0} \frac{1}{2} (\frac{π^{4} (2 + \cos (2 π z))}{3 \sin^{3} (π z)} - \frac{1}{z^{4}}) = \frac{π^{4}}{90}

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