Sequences and Series – Approximation Method for the Basel Problem

sequences-and-series

Basel Series

Suppose each term of the Basel series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ is represented as a rectangle of height $\frac{1}{n^2}$ and width $1$ along the x-axis.

The infinite sum of all those rectangle areas equals the Basel value of $\frac{\pi^2}{6}$.

basel

Approximation of Basel Series

Plot the curve $f(x) = \frac{1}{x^2}$

basel_approx

The area under the curve is

$$\int_{1}^{\infty} \frac{1}{x^2} \, dx = 1$$

The area of each rectangle is $\frac{1}{n^2}$, so the area above the curve is:

$$\frac{1}{n^2} – \int_{n}^{n+1} \frac{1}{x^2} \, dx$$

Calculating the areas of the portions of the rectangles above the curve $f(x) = \frac{1}{x^2}$ are

$\frac{1}{2}$, $\frac{1}{12}$, $\frac{1}{36}$, $\frac{1}{80}$, $\frac{1}{150}$, $\frac{1}{252}$, $\frac{1}{392}$, $\frac{1}{576}$, $\frac{1}{810}$, $\frac{1}{1100}, \cdots$

From OEIS, the denominator formula is

$$n^2 (n+1)$$

so the total sum of these areas above the curve is

$$\sum_{n=1}^{\infty} \frac{1}{n^2 (n+1)}$$

and adding the integral of the area below the curve gives

$$\overbrace{\int_{1}^{\infty} \frac{1}{x^2} \, dx}^{\text{area below curve}} + \underbrace{\sum_{n=1}^{\infty} \frac{1}{n^2 (n+1)}}_{\text{area above curve}}$$

Thus, the total infinite sum of all those rectangles is

$$1+\sum_{n=1}^{\infty} \frac{1}{n^2 (n+1)}$$

According to WolframAlpha, the result is

$$\frac{\pi^2}{6}$$

which is the exact value of the Basel Problem.

Question

Is this a valid method to approximate the Basel Problem?

Also, what steps did WolframAlpha take to evaluate the series

$$1+\sum_{n=1}^{\infty} \frac{1}{n^2 (n+1)}=\frac{\pi^2}{6}$$

wolframalpha

Best Answer

Also, what steps did WolframAlpha take to evaluate the series $$1+\sum_{n=1}^{\infty} \frac{1}{n^2 (n+1)}=\frac{\pi^2}{6}$$

See that $$\frac{1}{n^2 (n+1)} = \frac{n+1 - n}{n^2(n+1)} = \frac{1}{n^2} - \frac{1}{n(n+1)}. $$ The second term sums into $1$ by telescoping and most probably Wolframalpha knows that $\sum \frac1 {n^2} = \pi^2/6.$