The Basel Problem

Stack the terms 1n2\frac{1}{n^2} as 3D bars and watch the staircase of partial sums approach the plane at π26\frac{\pi^2}{6} without ever touching it. Why π\pi appears at all is only sketched at the idea level.

∑n=1∞1n2=π26\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}

Examples & Exercises

Prove the tail bound RN=∑n>N1n2<1NR_N = \sum_{n>N}\frac{1}{n^2} < \frac{1}{N}.

RN<∑n>N(1n−1−1n)=1NR_N < \sum_{n>N}\left(\frac{1}{n-1}-\frac{1}{n}\right) = \frac{1}{N}
N=20N = 20: the actual tail ≈0.0488<1/20=0.05\approx 0.0488 < 1/20 = 0.05.
Show proof
  1. For n≥2n \ge 2, 1n2<1n(n−1)=1n−1−1n\frac{1}{n^2} < \frac{1}{n(n-1)} = \frac{1}{n-1}-\frac{1}{n}.
  2. The finite sum telescopes: ∑n=N+1M(1n−1−1n)=1N−1M<1N\sum_{n=N+1}^{M}\left(\frac{1}{n-1}-\frac{1}{n}\right) = \frac{1}{N}-\frac{1}{M} < \frac{1}{N}. Letting M→∞M \to \infty gives RN<1NR_N < \frac{1}{N} (strict, since each term comparison is strict).
  3. For example with N=20N = 20 the tail is below 0.050.05 — the partial sum S20S_{20} is already within 0.050.05 of the limit.

Derive ∑n=1∞1(2n)2=π224\sum_{n=1}^{\infty}\frac{1}{(2n)^2} = \frac{\pi^2}{24} from the Basel sum.

∑n=1∞1(2n)2=14∑n=1∞1n2=π224\sum_{n=1}^{\infty}\frac{1}{(2n)^2}=\frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{24}
The even terms alone converge the same way.
Show proof
  1. 1(2n)2=14⋅1n2\frac{1}{(2n)^2} = \frac{1}{4}\cdot\frac{1}{n^2}.
  2. For a positive-term series constants factor out, so ∑n=1∞1(2n)2=14∑n=1∞1n2=14⋅π26=π224\sum_{n=1}^{\infty}\frac{1}{(2n)^2} = \frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{1}{4}\cdot\frac{\pi^2}{6} = \frac{\pi^2}{24}.

Derive ∑k=0∞1(2k+1)2=π28\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2} = \frac{\pi^2}{8}.

∑odd1n2=π26−π224=π28\sum_{\mathrm{odd}}\frac{1}{n^2}=\frac{\pi^2}{6}-\frac{\pi^2}{24}=\frac{\pi^2}{8}
Subtracting the even terms leaves exactly the odd ones.
Show proof
  1. A positive-term series splits into (even terms) + (odd terms).
  2. Hence the odd sum =π26−π224=4π2−π224=3π224=π28= \frac{\pi^2}{6} - \frac{\pi^2}{24} = \frac{4\pi^2-\pi^2}{24} = \frac{3\pi^2}{24} = \frac{\pi^2}{8}.

6. Series