The Basel Problem

A staircase of reciprocal squares climbs toward one flat plane. The height of that plane is the famous answer.

Each new step of the partial sums is smaller than the one before. So the staircase piles up toward a definite height. The surprise is that Ο€ hides inside that height. Why Ο€ appears at all is only sketched at the idea level.

For example, stack πŸ“¦ boxes under a line drawn on a wall. The boxes shrink until the stack creeps so close to the line that your eye cannot see the gap.

βˆ‘n=1∞1n2=Ο€26\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}

Examples & Exercises

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

RN<βˆ‘n>N(1nβˆ’1βˆ’1n)=1NR_N < \sum_{n>N}\left(\frac{1}{n-1}-\frac{1}{n}\right) = \frac{1}{N}
N=20N = 20: the actual tail β‰ˆ0.0488<1/20=0.05\approx 0.0488 < 1/20 = 0.05.
Show proof

  1. For nβ‰₯2n \ge 2, 1n2<1n(nβˆ’1)=1nβˆ’1βˆ’1n\frac{1}{n^2} < \frac{1}{n(n-1)} = \frac{1}{n-1}-\frac{1}{n}.

  2. The finite sum telescopes: βˆ‘n=N+1M(1nβˆ’1βˆ’1n)=1Nβˆ’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 Mβ†’βˆžM \to \infty alone gives only RN≀1NR_N \le \frac{1}{N}.

    To restore strictness, note the difference: 1n(nβˆ’1)βˆ’1n2=1n2(nβˆ’1)>0\frac{1}{n(n-1)} - \frac{1}{n^2} = \frac{1}{n^2(n-1)} > 0, so for every MM the difference sum is at least its first term: βˆ‘n=N+1M(1n(nβˆ’1)βˆ’1n2)β‰₯1N(N+1)2>0\sum_{n=N+1}^{M}\left(\frac{1}{n(n-1)} - \frac{1}{n^2}\right) \ge \frac{1}{N(N+1)^2} > 0.

    Hence βˆ‘n=N+1M1n2≀1Nβˆ’1Mβˆ’1N(N+1)2\sum_{n=N+1}^{M}\frac{1}{n^2} \le \frac{1}{N} - \frac{1}{M} - \frac{1}{N(N+1)^2}, and letting Mβ†’βˆžM \to \infty gives RN≀1Nβˆ’1N(N+1)2<1NR_N \le \frac{1}{N} - \frac{1}{N(N+1)^2} < \frac{1}{N}.

  3. For example with N=20N = 20 the tail is below 0.050.05 β€” the partial sum S20S_{20} is already within 0.050.05 of the limit.

Derive βˆ‘n=1∞1(2n)2=Ο€224\sum_{n=1}^{\infty}\frac{1}{(2n)^2} = \frac{\pi^2}{24} from the Basel sum.

βˆ‘n=1∞1(2n)2=14βˆ‘n=1∞1n2=Ο€224\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=14β‹…1n2\frac{1}{(2n)^2} = \frac{1}{4}\cdot\frac{1}{n^2}.

  2. For a positive-term series constants factor out, so βˆ‘n=1∞1(2n)2=14βˆ‘n=1∞1n2=14β‹…Ο€26=Ο€224\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 βˆ‘k=0∞1(2k+1)2=Ο€28\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2} = \frac{\pi^2}{8}.

βˆ‘odd1n2=Ο€26βˆ’Ο€224=Ο€28\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 =Ο€26βˆ’Ο€224=4Ο€2βˆ’Ο€224=3Ο€224=Ο€28= \frac{\pi^2}{6} - \frac{\pi^2}{24} = \frac{4\pi^2-\pi^2}{24} = \frac{3\pi^2}{24} = \frac{\pi^2}{8}.

7. Series