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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

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    $\begingroup$ By a partial fraction expansion, your new series telescopes, so yes, it's identical to the original Basel problem series. And your geometric argument with the integral is sound. But this isn't enough to evaluate the series, since you're not very far from what you had to begin with. You could however extend the integral comparison further with the Euler Maclaurin formula which is one known way to compute it. $\endgroup$
    – Jam
    Commented Dec 14, 2023 at 16:10

1 Answer 1

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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 $\sum \frac1 {n^2} = \pi^2/6.$

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  • $\begingroup$ Thanks...and I was kind of hoping that the basel value didn't appear in the evaluation. Can the series be evaluated by excluding the known basel value? That would be nice i.e., $$\sum_{n=1}^{\infty} \frac{1}{n^2} = 1+\sum_{n=1}^{\infty} \frac{1}{n^2 (n+1)} = \frac{\pi^2}{6}$$ $\endgroup$
    – vengy
    Commented Dec 14, 2023 at 19:40
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    $\begingroup$ @vengy The two series are equal, so they will both be solved to the same thing :) $\endgroup$
    – Jam
    Commented Dec 14, 2023 at 20:36
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    $\begingroup$ Thanks. Looks like the general formula is $$\frac{1}{s - 1} + \sum_{n=1}^{\infty} \left( \frac{1}{n^s} + \frac{1}{(s - 1)(n + 1)^{s - 1}} - \frac{1}{(s - 1)n^{s - 1}} \right) = \zeta(s)$$ $\endgroup$
    – vengy
    Commented Dec 23, 2023 at 2:03

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