ζ(4) and Even Powers | Real Analysis

The partial sums hug the answer after only a few terms. Fourth powers shrink the tail to almost nothing. At every even power, the closed answer is a power of pi times a fraction.

For example, picture stacking 🍰 cake slices that shrink fast. Each slice is soon tiny, so a few slices fix the total height.

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.)
$\frac{15}{16}\cdot\frac{\pi^4}{90} = \frac{\pi^4}{96}$ .

For $N \ge 1$ , 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

Since $f(x) = x^{-4}$ is strictly decreasing, for $n \ge 2$ and $x \in [n-1, n)$ we have $x^{-4} > n^{-4}$ , so the inequality is strict: $\frac{1}{n^4} < \int_{n-1}^{n}\frac{dx}{x^4}$ .
For $N \ge 1$ the tail starts at $n = N+1 \ge 2$ , so the bound above applies to every tail term, giving the strict bound $R_N < \sum_{n>N}\int_{n-1}^{n}\frac{dx}{x^4} = \int_{N}^{\infty}\frac{dx}{x^4} = \frac{1}{3N^3}$ .
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 $\zeta(2n) = \frac{(2\pi)^{2n}|B_{2n}|}{2(2n)!}$ gives all three values exactly.
For $n = 1$ : $\frac{(2\pi)^2(1/6)}{2\cdot 2!} = \frac{4\pi^2/6}{4} = \frac{\pi^2}{6}$ .
For $n = 2$ : $\frac{(-1)^3 B_4 (2\pi)^4}{2\cdot 4!} = \frac{(1/30)\cdot 16\pi^4}{48} = \frac{\pi^4}{90}$ .
For $n = 3$ : $\frac{(2\pi)^6(1/42)}{2\cdot 6!} = \frac{64\pi^6}{42\cdot 1440} = \frac{\pi^6}{945}$ .
$\zeta(3)$ (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.