An “original” proof of the Basel Sum

I recently got bored and decided to try to find a proof of the Basel Sum. In an unexpected departure from tradition, I actually found one. As a further surprise, it seems to be a lot simpler than any other proof I’ve seen: just a few steps, and each one pretty obvious.

A reminder: the Basel sum states that

    \begin{equation*} \sum_{1}^{\infty}\frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \ldots = \pi^2/6 \end{equation*}

And an important clarification: this proof is “original”, in the sense of I came up with it myself, yay, but most definitely not original in the sense of I was the first to come up with this proof. The Basel sum is exceedingly well traveled territory mathematically, and I would be surprised if I was in the first 1000 people to come up with this proof independently.

Before we start the proof, a reminder of the definition of the double factorial. For an even number 2k, 2k!! = 2 \times 4 \times 6 \ldots \times (2k). Similarly, for an odd number (2k+1), (2k+1)!! = 1 \times 3 \times 5 \ldots \times (2k+1). And we can choose to define (-1)!! = 0!! = 1.

Step 1:

Let

    \[S = \sum_{1}^{\infty} \frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \ldots\]

Then,

    \[\frac{S}{4} = \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \ldots\]

Subtracting these two equations, we get

(1)   \begin{equation*} \boxed{\frac{3S}{4} = \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots} \end{equation*}

Step 2:

(2)   \begin{equation*} \boxed{\int_{0}^{\pi/2} \sin^{2k+1}(\theta) dx = \frac{(2k)!!}{(2k+1)!!}} \end{equation*}

This is easy to prove by induction, using integration by parts.
Click here for the proof, if you really, really want it. But you probably don't.

Seriously, no feelings will be hurt if you skip this.

Define

    \[I_{2k+1} = \int_{0}^{\pi/2} \sin^{2k+1}(\theta) d\theta\]

Then,

    \begin{equation*} \begin{split} I_{2k+1} & = -\sin^{2k}\theta \cos\theta \bigg\rvert_{0}^{\pi/2} -\int_{0}^{\pi/2} (-\cos^2\theta) (2k) \sin^{2k-1}\theta d\theta \\ & = (0) + (2k) \int_{0}^{\pi/2} (1 - \sin^2(\theta) \sin^{2k-1}(\theta) d\theta \\ & \Rightarrow I_{2k+1} = 2k I_{2k-1} - 2k I_{2k+1} \\ & \Rightarrow I_{2k+1} = \frac{2k}{2k+1} I_{2k-1} \\ & \Rightarrow I_{2k+1} = \frac{(2k)!!}{(2k+1)!!} \\ \end{split} \end{equation*}

Hope you feel it was worth it!

Step 3:
Start with the Taylor series for \arcsin(x):

    \[\arcsin(x) = x + \frac{1}{2} \frac{x^3}{3} + \frac{1\cdot 3}{2\cdot 4} \frac{x^5}{5} + \ldots = \sum_{k=0}^{\infty} \frac{(2k-1)!!}{(2k)!!} \frac{x^{2k+1}}{2k+1}\]

Note that this is absolutely convergent. Now simply substitute x = \sin \theta.

(3)   \begin{equation*} \boxed{\theta = \sum_{k=0}^{\infty} \frac{(2k-1)!!}{(2k)!!} \frac{\sin(\theta)^{2k+1}}{2k+1}} \end{equation*}

Step 4:
In equation (3), integrate from 0 to \pi/2.

    \begin{equation*} \begin{split} LHS & = \int_{0}^{\pi/2} \theta d\theta \\ & = \pi^2/8 \\ RHS & = \sum_{0}^{\infty} \frac{(2k-1)!!}{(2k)!!} \frac{I_{2k+1}}{2k+1} \\ & = \sum_{0}^{\infty} \frac{(2k-1)!!}{(2k)!!} \frac{1}{2k+1} \frac{(2k)!!}{(2k+1)!!} \\ & = \sum_{0}^{\infty} \frac{1}{(2k+1)^2} \end{split} \end{equation*}

Step 5:
So the RHS is what we looked at in equation (1). This means

    \[\frac{3S}{4} = \pi^2/8\]

so

(4)   \begin{equation*} \boxed{\sum_{1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}} \end{equation*}

Addendum: I did some more research. Euler himself did something very similar (here). The earliest version of this exact proof that I could find was by Boo Rim Choe in the American Mathematical Monthly in 1987 (Vol. 94, No. 7 (Aug. – Sep., 1987), pp. 662-663 ).

One thought on “An “original” proof of the Basel Sum”

  1. Pingback: A pretty straightforward proof of the Basel Sum - Nevin Manimala's Blog

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