Basel Problem Proof

If we define $S_n \; =\; \frac{1}{2^{2n+4}}\sum_{k=1}^{2^n} \mathrm{cosec}^2\,\left(\tfrac{(2k-1)\pi}{2^{n+2}}\right)$ for any integer $n \ge 0$ then $S_0 = \tfrac{1}{16}\mathrm{cosec}^2\,\tfrac14\pi = \tfrac18$ . Note also that $S_n \; = \; \frac{1}{2^{2n+4}}\sum_{k=1}^{2^n} \mathrm{cosec}^2\,\left(\tfrac{(2k-1)\pi}{2^{n+2}}\right) \; = \; \frac{1}{2^{2n+4}}\sum_{k=1}^{2^n} \mathrm{cosec}^2\,\left(\tfrac{(2(2^n+1-k)-1)\pi}{2^{n+2}}\right) \; = \; \frac{1}{2^{2n+4}}\sum_{k=1}^{2^n} \sec^2\,\left(\tfrac{(2k-1)\pi}{2^{n+2}}\right)$ and so $\begin{aligned} S_{n+1} & = \; \frac{1}{2^{2n+7}}\sum_{k=1}^{2^{n+1}} \left\{\mathrm{cosec}^2\,\left(\tfrac{(2k-1)\pi}{2^{n+3}}\right) + \sec^2\,\left(\tfrac{(2k-1)\pi}{2^{n+3}}\right)\right\} \; = \; \frac{1}{2^{2n+7}}\sum_{k=1}^{2^{n+1}} \mathrm{cosec}^2\,\left(\tfrac{(2k-1)\pi}{2^{n+3}}\right)\sec^2\,\left(\tfrac{(2k-1)\pi}{2^{n+3}}\right) \\ & = \; \frac{1}{2^{2n+5}}\sum_{k=1}^{2^{n+1}} \mathrm{cosec}^2\,\left(\tfrac{(2k-1)\pi}{2^{n+2}}\right) \; = \; \frac{1}{2^{2n+5}}\sum_{k=1}^{2^{n}} \left\{\mathrm{cosec}^2\,\left(\tfrac{(2k-1)\pi}{2^{n+2}}\right) + \mathrm{cosec}^2\,\left(\tfrac{(2(k+2^n)-1)\pi}{2^{n+2}}\right)\right\} \\ & = \; \frac{1}{2^{2n+5}}\sum_{k=1}^{2^{n}} \left\{\mathrm{cosec}^2\,\left(\tfrac{(2k-1)\pi}{2^{n+2}}\right) + \sec^2\,\left(\tfrac{(2k-1)\pi}{2^{n+2}}\right)\right\} \; = \; \tfrac12S_n + \tfrac12S_n \; =\; S_n \end{aligned}$ and hence $S_n = \tfrac18$ for all positive integers $n$ .

We will use induction on n. The base case of n=0 is straghtforward. For the iterative step, we will show that the k'th term out of $2^{n}$ is equal to the k'th term out of $2^{n+1}$ plus the $2^{n+1}+1-k$ 'th term out of $2^{n+1}$ . In equation form, we must show that $\frac{1}{2^{2n+4} \sin^2 \left(\frac{2\pi k-\pi}{2^{n+2}}\right)}=\frac{1}{2^{2n+6} \sin^2 \left(\frac{2\pi (2^{n+1}+1-k)-\pi}{2^{n+3}}\right)}+\frac{1}{2^{2n+6} \sin^2 \left(\frac{2\pi k-\pi}{2^{n+3}}\right)}$
$\frac{1}{2^{2n+6} \sin^2 \left(\frac{2\pi (2^{n+1}+1-k)-\pi}{2^{n+3}}\right)}+\frac{1}{2^{2n+6} \sin^2 \left(\frac{2\pi k-\pi}{2^{n+3}}\right)}=\frac{1}{2^{2n+6} \cos^2 \left(\frac{2\pi k-\pi}{2^{n+3}}\right)}+\frac{1}{2^{2n+6} \sin^2 \left(\frac{2\pi k-\pi}{2^{n+3}}\right)}=\frac{1}{2^{2n+6} \cos^2 \left(\frac{2\pi k-\pi}{2^{n+3}}\right) \sin^2 \left(\frac{2\pi k-\pi}{2^{n+3}}\right)}$ Using the double angle identity for sin(x), we can show that $\frac{1}{2^{2n+4} \sin^2 \left(\frac{2\pi k-\pi}{2^{n+2}}\right)}=\frac{1}{2^{2n+6} \cos^2 \left(\frac{2\pi k-\pi}{2^{n+3}}\right) \sin^2 \left(\frac{2\pi k-\pi}{2^{n+3}}\right)}$ which gives us the desired result. By the way, this property is not specific to powers of 2. If we substitute n for $2^{n}$ , the equality remains true; powers of 2 are just a lot easier to prove than the natural numbers.

Bonus: Because sum is equal to 1/8 for all n, we will consider the limit of each term in the sum as n goes to infinity. Using the Taylor series for sin(x), we obtain $\frac{1}{\pi^{2} (2k-1)^{2}+e}$ where e is a small error whose most significant term is proportional to $2^{-2n}$ . Notice that the amount of error shrinks more quickly than the number of terms ( $2^n$ ) that the error appears in grows. Therefore, since the limit as n goes to infinity of each individual term is $\frac{1}{\pi^{2} (2k-1)^{2}}$ , the limit of the entire sum as n goes to infinity is $\sum_{k=1}^\infty \big(\frac{1}{\pi^{2} (2k-1)^{2}}\big)$ Let the sum of the reciprocals of the squares (i.e. the Basel Problem) be S. Notice that S/4+ $\pi^{2}$ *(the infinite sum that we just created)=S. Substituting 1/8 for (the infinite sum that we just created) shows us that the sum of the reciprocals of the squares is equal to $\frac{\pi^{2}}{6}$ .

Note: Some of you may be wondering why our error term had to shrink at a greater rate than the rate at which that the number of terms in the sum grew. Consider the sum $\sum_{k=1}^{n^2} \big(\frac{1}{k^2}+\frac{1}{n}\big)$ Even though the limit as n goes to infinity of each individual term is $1/k^2$ , the error term of 1/n gets added $n^2$ times, so the limit of the entire sum is infinity, not $\frac{\pi^2}{6}$ . Similarly, $\frac{1}{\pi^{2} (2k-1)^{2}+e}$ can be rearranged to show that not only does the error term tend to 0 for large n, but the sum of the differences( $\frac{1}{\pi^{2} (2k-1)^{2}+e}-\frac{1}{\pi^{2} (2k-1)^{2}}$ ) caused by all $2^n$ of the error terms tends to 0 also.