Find the sum of 1/n^2 for n from 1 to infinity.

Step-by-step solution

1. 1

   Confirm convergence before hunting for a value. This is a $p$-series $\sum n^{-p}$ with $p=2$. The integral test compares it with $\int_{1}^{\infty}x^{-2}dx=1$, which is finite, so the series converges — unlike the harmonic series $p=1$, which diverges. Asking for a sum is only meaningful once this is settled.

2. 2

   Bound the sum crudely to know what to expect. Since $\dfrac{1}{n^{2}}<\dfrac{1}{n(n-1)}$ for $n\ge 2$, and that telescopes to $1$, the tail after the first term is under $1$:

   $1<\sum_{n=1}^{\infty}\frac{1}{n^{2}}<2$

   So the answer lies between 1 and 2 — a useful guard against a wrong closed form.

3. 3

   Recall Euler idea: factor $\dfrac{\sin x}{x}$ over its roots. The function $\dfrac{\sin x}{x}$ equals $1$ at $x=0$ and vanishes at $x=\pm k\pi$, which suggests the infinite product

   $\frac{\sin x}{x}=\prod_{k=1}^{\infty}\left(1-\frac{x^{2}}{k^{2}\pi^{2}}\right)$

   This treats a transcendental function like a polynomial written from its roots, the leap that made the argument famous.

4. 4

   Match the $x^{2}$ coefficients. Expanding the product, the coefficient of $x^{2}$ is $-\displaystyle\sum_{k\ge 1}\frac{1}{k^{2}\pi^{2}}$. From the Taylor series $\dfrac{\sin x}{x}=1-\dfrac{x^{2}}{6}+\cdots$, that same coefficient is $-\dfrac16$. Equating:

   $\frac{1}{\pi^{2}}\sum_{k=1}^{\infty}\frac{1}{k^{2}}=\frac16$

5. 5

   Solve for the sum.

   $\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}\approx 1.6449340668$

   It sits comfortably inside the bounds from step 2. A direct partial sum of two million terms gives $1.6449335668$, and the missing $5\times 10^{-7}$ is exactly the tail $\approx 1/N$ ✓. In modern notation this is $\zeta(2)$.