ζ(4) and Even Powers | Real Analysis

The partial sums of

\frac{1}{n^4}

reach

\frac{\pi^4}{90}

quickly — and at every even power, the

\zeta

value closes into a power of

\pi

Examples & Exercises

Derive $\sum_{k=0}^{\infty}\frac{1}{(2k+1)^4} = \frac{\pi^4}{96}$ from $\zeta(4)$ .

\sum_{\mathrm{odd}}\frac{1}{n^4}=\zeta(4)-\frac{1}{16}\zeta(4)=\frac{15}{16}\cdot\frac{\pi^4}{90}=\frac{\pi^4}{96}

Removing the even terms leaves exactly the odd ones.

Show proof

Even terms: $\sum_{n=1}^{\infty}\frac{1}{(2n)^4} = \frac{1}{16}\sum_{n=1}^{\infty}\frac{1}{n^4} = \frac{1}{16}\zeta(4)$ .
Odd terms = whole − even $= \zeta(4)-\frac{1}{16}\zeta(4) = \frac{15}{16}\zeta(4)$ . (Rearranging a positive-term series is legitimate.)
$\frac{15}{16}\cdot\frac{\pi^4}{90} = \frac{\pi^4}{96}$ .

Prove $R_N = \sum_{n>N}\frac{1}{n^4} < \frac{1}{3N^3}$ by integral comparison.

R_N < \int_{N}^{\infty}\frac{dx}{x^4} = \frac{1}{3N^3}

N = 8

: actual tail

\approx 0.000539 < 1/1536 \approx 0.000651

Show proof

$f(x) = x^{-4}$ is decreasing, so for $n \ge 2$ , $\frac{1}{n^4} \le \int_{n-1}^{n}\frac{dx}{x^4}$ .
Hence $R_N \le \sum_{n>N}\int_{n-1}^{n}\frac{dx}{x^4} = \int_{N}^{\infty}\frac{dx}{x^4} = \frac{1}{3N^3}$ .
For $N = 8$ the bound is $\frac{1}{1536} \approx 0.000651$ — consistent with the actual tail $0.000539$ .

Check $\zeta(2)=\frac{\pi^2}{6}$ , $\zeta(4)=\frac{\pi^4}{90}$ , $\zeta(6)=\frac{\pi^6}{945}$ against the Bernoulli-number formula. (The formula itself is taken as known.)

\zeta(2n) = \frac{(-1)^{n+1}B_{2n}(2\pi)^{2n}}{2(2n)!}

Every even zeta value is a power of

\pi

over an integer.

Show proof

Substituting $B_2 = \frac{1}{6}$ , $B_4 = -\frac{1}{30}$ , $B_6 = \frac{1}{42}$ into the formula gives all three values exactly. For $n = 2$ : $\frac{(-1)^3 B_4 (2\pi)^4}{2\cdot 4!} = \frac{(1/30)\cdot 16\pi^4}{48} = \frac{\pi^4}{90}$ .
$\zeta(3)$ (Apéry's constant) has no such form — the pattern is for even arguments only.
The proof of the formula is not covered here — it is taken as a known result.