Basel Problem:Using Resolution Into Factors

Author

Abhirup Moitra

Published

July 5, 2023

Leonard Euler’s calculational prowess was remarkable, including his work in the area of infinite series. His methods are not covered in this article, but we will state and prove one of his famous results. We know the series converges and we’ll show how to prove that. But the problem of determining its exact value is quite difficult. The problem was known as the Basel problem, named after Basel University in Switzerland, and Euler solved it in 1735 when he obtained a surprising result that \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{{\pi}^2}{6}.\) He also described a process to derive the values of the series with higher even power of \(\displaystyle\frac{1}{n}.\) For example, he showed that \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{{\pi}^4}{90}\) and \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^6} = \frac{{\pi}^6}{945}.\) However, the values of the corresponding series of odd powers of \(\displaystyle\frac{1}{n}\)were not discovered and stay elusive to the present day.

Convergence of \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}\)

Approach 1

\(\displaystyle\text{Observe that}\; \sum_{n=1}^{\infty} \frac{1}{n^2}\; \text{convergent iff}\; \displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2}\; \text{convergent.}\)

\(\text{Note that;}\; 0\le \frac{1}{n^2}\le \frac{1}{n^2-n}=\frac{1}{n-1}-\frac{1}{n}\)

\(\text{From the Comparison test, we know that if}\ \{x_n\}_{n\ge1}\ \text{and}\) \(\{y_n\}_{n\ge1}\; \text{be real sequences}\) \(\text{and for some}\; k\in \mathbb{N}\; \text{we have,} \; 0\le x_n \le y_n\; \forall n\ge k\; \text{then,}\)

\(\text{The convergence of}\; \sum y_n\; \text{implies the convergence of}\; \sum x_n.\)

\(\text{By comparison test,}\) \(\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2-n} \; \text{convergent implies}\; \sum_{n=2}^{\infty} \frac{1}{n^2}\; \text{converges.}\)

\(y_n = \frac{1}{n^2-n}= \frac{1}{n-1}-\frac{1}{n}\)

\(\displaystyle\lim_{N \to \infty} \; \sum_{n=2}^{\infty}\; y_n\)

\(= \displaystyle\lim_{N \to \infty} \; \sum_{n=2}^{\infty}\; \big( \frac{1}{n-1}-\frac{1}{n}\big)\)

\(=\displaystyle\lim_{N \to \infty} \big[\big(\frac{1}{1}-\frac{1}{2}\big)+ \big(\frac{1}{2}-\frac{1}{3}\big)+\big(\frac{1}{3}-\frac{1}{4}\big)+\ldots+\big(\frac{1}{N}-\frac{1}{N-1}\big)\big]\)

\(= \displaystyle\lim_{N \to \infty} \; \big[1-\frac{1}{N}\big] = 1\)

\(\therefore \; \displaystyle\sum_{n=2}^{\infty} y_n = \sum_{n=2}^{\infty} \frac{1}{n^2-n} \; \text{converges to} \; 1. \;\text{Hence} \sum_{n=2}^{\infty} \;\frac{1}{n^2} \text{is convergent.}\)

\(\hspace{6 cm}\boxed {\therefore \sum_{n=1}^{\infty} \frac{1}{n^2} \text{is convergent.}}\)

Approach 2

Since some partial sums are monotone, it suffices to show that some sub-sequences of \(\{s_k\}\) is bounded. If \(k_1:=2^1 - 1=1,\) then \(s_{k_{1}} = 1.\) If \(k_2:= 2^2-1 = 3,\) then

\[s_{k_{2}} = \frac{1}{1}+\bigg(\frac{1}{2^2}+\frac{1}{3^2}\bigg)<1+\frac{2}{2^2}=1+ \frac{1}{2}\]

and if \(k_3:= 2^3-1=7,\) then we have

\[ s_{k_3}= s_{k_2}+ \bigg(\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}\bigg)< s_{k_2}+\frac{4}{4^2}<1+\frac{1}{2}+\frac{2}{2^2} \]

By mathematical induction, we find that if \(k_j:=2^j-1,\) then

\[ 0<s_{k_j}<1+\frac{1}{2}+\bigg(\frac{1}{2}\bigg)^2+\ldots+\bigg(\frac{1}{2}\bigg)^{j-1} \]

Since the term on the right is a partial sum of a geometric series with \(r=\frac{1}{2}\), it is dominated by \(\frac{1}{1-\frac{1}{2}}=2\). But we know that, if \(\{x_n\}\) is a sequence of non-negative real numbers. Then the series \(\sum x_n\) converges iff the sequence \(\{s_k\}\) of partial sums are bounded. In this case,

\[ \boxed{\sum_{n=1}^{\infty}x_n=\lim (s_k)= \text{sup}\{s_k:k\in \mathbb{N}\}} \]

This implies that \(\sum_{n=1}^{\infty} \frac{1}{n^2}\) is convergent.

Resolution Into Factors

Resolve \(\sin{\theta}\) Into Factors

Here our goal is to express \(\sin\theta\) as an infinite product i.e.,

\[ \sin \theta =\theta\prod_{r=1}^{\infty} \bigg(1- \dfrac{{\theta}^2}{r^2{\pi}^2}\bigg) = \theta\bigg(1- \dfrac{{\theta}^2}{{\pi}^2}\bigg) \bigg(1- \dfrac{{\theta}^2}{2^2{\pi}^2}\bigg) \bigg(1- \dfrac{{\theta}^2}{3^2{\pi}^2}\bigg) \ldots \infty \]

Proof

We have,

\[ \sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}= 2 \sin \frac{A}{2} \sin \frac{\pi+A}{2} \; \hspace{1.7 cm}\ldots (1) \]

then,

\[ \hspace{2.5 cm}\boxed{\sin \theta = 2 \sin \frac{\theta}{2} \sin \frac{\pi+\theta}{2}}\; \hspace{2.1 cm}\ldots (2) \]

Substituting \(\frac{\theta}{2}\) and \(\frac{\pi+\theta}{2}\) for \(A\) successively in the formula \((1),\) we get,

\[ \boxed{\sin \frac{\theta}{2} = 2 \sin \frac{{\theta}}{2^2} \sin \frac{2 \pi+\theta}{2^2}} \]

and

\[ \boxed{\sin \frac{\pi+\theta}{{2}} = 2\sin \frac{\pi+\theta}{{2^2}} \; \sin \frac{3\pi+\theta}{{2^2}}} \]

Putting these values in the right side of \((2)\) and re-arranging them, we have

\[ \hspace{1.8 cm}\boxed{\sin \theta = 2^3 \sin \frac{\theta}{2^2} \sin \frac{\pi+\theta}{2^2} \sin \frac{2\pi+\theta}{{2}^2} \sin \frac{3\pi+\theta}{{2}^2}} \hspace{2 cm}\ldots (3) \]

Similarly, substituiting \(\frac{{\theta}}{2^2}, \frac{\pi+\theta}{2^2},\frac{2\pi+\theta}{2^2}, \frac{3\pi+\theta}{2^2}\) for \(A\) successively in \((1)\) and putting these values in the right side of \((3)\) and re-arranging them, we get

\[\sin\theta = 2^7 \sin \frac{\theta}{2^3} \sin \frac{\pi+\theta}{{2^3}}\sin \frac{2\pi+\theta}{{2}^3}\ldots \;\sin \frac{7\pi+\theta}{{2}^3}\] \[\implies \sin \theta = 2^{2^3-1} \sin \frac{\theta}{2^3} \sin \frac{\pi+\theta}{{2^3}}\sin \frac{2\pi+\theta}{{2}^3} \ldots \sin \frac{(2^3-1)\pi+\theta}{2^3}\]

\[ \hspace{1.9 cm} = \ldots \ldots \ldots \ldots \ldots..................................... \]

\[ \hspace{1.9 cm} = \ldots \ldots \ldots \ldots \ldots………………………………. \]

\[ \hspace{1.9 cm} = 2^{p-1} \sin \frac{\theta}{p} \sin \frac{\pi+\theta}{{p}}\sin \frac{2\pi+\theta}{p} \ldots \sin \frac{(p-1)\pi+\theta}{p} \]

Therefore,

\[ \hspace{1.7 cm}\boxed{\sin\theta = 2^{p-1} \sin \frac{\theta}{p} \sin \frac{\pi+\theta}{{p}}\sin \frac{2\pi+\theta}{p} \ldots \sin \frac{(p-1)\pi+\theta}{p}} \hspace{1.5 cm} \ldots (4) \]

where \(p = 2^n.\) The last factor in \((4)\) is \(\sin \frac{(p-1)\pi+\theta}{p}\)

\[ \sin \frac{(p-1)\pi+\theta}{p} = \sin\bigg(\pi-\dfrac{\pi-\theta}{p}\bigg) = \sin\bigg(\dfrac{\pi-\theta}{p}\bigg) \]

The last but one factor in \((4)\) is \(\sin \frac{(p-2)\pi+\theta}{p}\)

\[ \sin \frac{(p-2)\pi+\theta}{p} = \sin\bigg(\pi-\dfrac{2\pi-\theta}{p}\bigg) = \sin\bigg(\dfrac{2\pi-\theta}{p}\bigg) \]

, and so on. The \({(\frac{1}{2}p+1)}^{\text{th}}\) factor from the beginning

\[\sin\Biggl(\dfrac{\frac{1}{2}p\pi+\theta}{p}\Biggl) = \sin\Biggl(\dfrac{\pi}{2}+\dfrac{\theta}{p}\Biggl) = \cos \dfrac{\theta}{p}\]Hence combining the second and the last factors, the third and the last but one, and so on, and leaving alone the first and \({(\frac{1}{2}p+1)}^{\text{th}}\) factors which have no conjugates, the equation \((4)\) becomes

\[ \boxed{\sin\theta = 2^{p-1} \sin \frac{\theta}{p}\Biggl\{\sin^2\dfrac{\pi}{p}- \sin^2 \dfrac{\theta}{p} \Biggl\} \Biggl\{\sin^2\dfrac{2\pi}{p}- \sin^2 \dfrac{\theta}{p} \Biggl\}\ldots \Biggl\{\sin^2\dfrac{(\frac{1}{2}p-1)\pi}{p}- \sin^2 \dfrac{\theta}{p} \Biggl\}\ \cos \dfrac{\theta}{p}} \]

\(\hspace{14.7 cm} \ldots (5)\)

\[ \implies \boxed{\dfrac{\sin \theta}{\sin \frac{\theta}{p}} = \Biggl\{\sin^2\dfrac{\pi}{p}- \sin^2 \dfrac{\theta}{p} \Biggl\} \Biggl\{\sin^2\dfrac{2\pi}{p}- \sin^2 \dfrac{\theta}{p} \Biggl\}\ldots \Biggl\{\sin^2\dfrac{(\frac{1}{2}p-1)\pi}{p}- \sin^2 \dfrac{\theta}{p} \Biggl\}\ \cos \dfrac{\theta}{p}} \]

\(\hspace{14.8 cm} \ldots (6)\)

Let \(\theta \to 0,\) then \(\displaystyle\lim_{\theta \to 0}\Biggl(\dfrac{\frac{\sin\theta}{\theta}}{\frac{\sin \frac{\theta}{p}}{\frac{\theta}{p}}}\Biggl)p=p, \; \lim_{\theta \to 0} \sin \frac{\theta}{p} = 0 \;\ \text{and}\; \lim_{\theta \to 0} \cos \frac{\theta}{p} = 1.\) Thus equation \((6)\) becomes

\[ \hspace{2.5 cm}\boxed{p = 2^{p-1} \sin^2 \dfrac{\pi}{p}\; \sin^2 \dfrac{2\pi}{p} \ldots \sin^2\dfrac{(\frac{1}{2}p-1)\pi}{p}} \; \hspace{2 cm} \ldots(7) \]

Dividing \((5)\) by \((7)\), we get

\[ \sin \theta = p \sin \frac{\theta}{p}\ \cos\dfrac{\theta}{p} \Biggr\{1- \dfrac{\sin^2 \frac{\theta}{p}}{\sin^2 \frac{\pi}{p}}\Biggr\} \Biggr\{1- \dfrac{\sin^2 \frac{\theta}{p}}{\sin^2 \frac{2\pi}{p}}\Biggr\} \ldots \]

Now, let \(p \to \infty\); then \(\displaystyle\lim_{p \to \infty} \bigg(p \sin \frac{\theta}{p}\bigg)= \lim_{p \to \infty} \Biggr(\dfrac{\sin \frac{\theta}{p}}{\frac{\theta}{p}}\Biggr).\theta = \theta\; ; \lim_{p \to \infty} \cos \frac{\theta}{p} =1\)

\[ \displaystyle \lim_{p \to \infty} \dfrac{\sin^2 \frac{\theta}{p}}{\sin^2 \frac{\pi}{p}} = \Biggr[\dfrac{\frac{\sin \frac{\theta}{p}}{\frac{\theta}{p}}}{\frac{\sin \frac{\pi}{p}}{\frac{\pi}{p}}}\Biggl]^2. \dfrac{{\theta}^2}{{\pi}^2} = \dfrac{{\theta}^2}{{\pi}^2} , \text{and so on.} \]

Hence,

\[ \boxed{\sin \theta = \theta\bigg(1- \dfrac{{\theta}^2}{{\pi}^2}\bigg) \bigg(1- \dfrac{{\theta}^2}{2^2{\pi}^2}\bigg) \bigg(1- \dfrac{{\theta}^2}{3^2{\pi}^2}\bigg) \ldots \infty} \]

i.e., \[\boxed{\sin \theta = \theta\prod_{r=1}^{\infty} \bigg(1- \dfrac{{\theta}^2}{r^2{\pi}^2}\bigg)}\hspace{2 cm} \text{(Proved)}\]

The Basel Problem

The Basel problem asks for the precise summation of the reciprocals of the squares of the natural numbers, i.e. the precise sum of the infinite series. The sum of the series is approximately equal to \(1.644934\).

Prove that, \(\displaystyle\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots \infty = \frac{{\pi}^2}{6}\) i.e., \(\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{{\pi}^2}{6}\)

\(\displaystyle \sin \theta = \theta \ - \frac{{\theta}^3}{3!}+\frac{{\theta}^3}{5!}-\ldots = \theta \Biggl(1- \frac{{\theta}^2}{6}+\frac{{\theta}^4}{120}-\ldots\Biggl)\)

We have also, \(\displaystyle \sin \theta = \theta\bigg(1- \dfrac{{\theta}^2}{{\pi}^2}\bigg) \bigg(1- \dfrac{{\theta}^2}{2^2{\pi}^2}\bigg) \bigg(1- \dfrac{{\theta}^2}{3^2{\pi}^2}\bigg) \ldots \infty\)

Therefore,

\(\displaystyle \theta \Biggl(1- \frac{{\theta}^2}{6}+\frac{{\theta}^4}{120}-\ldots\Biggl) = \theta\bigg(1- \dfrac{{\theta}^2}{{\pi}^2}\bigg) \bigg(1- \dfrac{{\theta}^2}{2^2{\pi}^2}\bigg) \bigg(1- \dfrac{{\theta}^2}{3^2{\pi}^2}\bigg) \ldots \infty\)

\(\displaystyle \implies 1- \Biggl(\frac{{\theta}^2}{6}-\frac{{\theta}^4}{120}+\ldots \infty\Biggl) = \bigg(1- \dfrac{{\theta}^2}{{\pi}^2}\bigg) \bigg(1- \dfrac{{\theta}^2}{2^2{\pi}^2}\bigg) \bigg(1- \dfrac{{\theta}^2}{3^2{\pi}^2}\bigg) \ldots \infty\)

Taking logarithms of both the sides, we get

\(\displaystyle \implies\ln \Biggl[ 1- \Biggl(\frac{{\theta}^2}{6}-\frac{{\theta}^4}{120}+\ldots \infty\Biggl)\Biggl] = \ln\bigg(1- \dfrac{{\theta}^2}{{\pi}^2}\bigg)+\ln \bigg(1- \dfrac{{\theta}^2}{2^2{\pi}^2}\bigg)+\ln \bigg(1- \dfrac{{\theta}^2}{3^2{\pi}^2}\bigg)+ \ldots \infty\)

\(\displaystyle \implies -\Biggl(\frac{{\theta}^2}{6}-\frac{{\theta}^4}{120}+\ldots \infty\Biggl)- \dfrac{1}{2} \Biggl(\frac{{\theta}^2}{6}-\frac{{\theta}^4}{120}+\ldots \infty\Biggl)^2 - \ldots\infty\)

\(=\displaystyle\Biggl(- \dfrac{{\theta}^2}{{\pi}^2}-\frac{1}{2}\dfrac{{\theta}^4}{{\pi}^4}-\ldots \infty\Biggl)+ \Biggl(-\dfrac{{\theta}^2}{2^2{\pi}^2}-\frac{1}{2}\dfrac{{\theta}^4}{2^4{\pi}^2}-\ldots\infty \Biggl)+ \Biggl(-\dfrac{{\theta}^2}{{3^2\pi}^2}-\frac{1}{2}\dfrac{{\theta}^4}{{3^4\pi}^4}-\ldots \infty\Biggl)\ldots \infty\)

\(\hspace{15 cm} \ldots (*)\)

Equating the co-efficients of \({\theta}^2\) from the both sides of this identity, we have

\[ \dfrac{1}{6} = \dfrac{{1}^2}{{\pi}^2}+\dfrac{{1}^2}{2^2{\pi}^2}+ \dfrac{{1}^2}{3{\pi}^2}+ \ldots \infty \]

\[ \implies \dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+ \ldots \infty = \dfrac{{\pi}^2}{6} \]

Therefore, \(\hspace{6.5 cm}\displaystyle \boxed{\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{{\pi}^2}{6}}\)

Illustration

From the L.H.S of the identity \((*)\) we get

\(\displaystyle -\Biggl(\frac{{\theta}^2}{6}-\frac{{\theta}^4}{120}+\ldots \infty\Biggl)- \dfrac{1}{2} \Biggl(\frac{{\theta}^2}{6}-\frac{{\theta}^4}{120}+\ldots \infty\Biggl)^2 - \ldots\infty\)

\(\displaystyle = -\Biggl(\frac{{\theta}^2}{6}-\frac{{\theta}^4}{120}+\ldots \infty\Biggl)\Biggl[1+ \dfrac{1}{2} \Biggl(\frac{{\theta}^2}{6}-\frac{{\theta}^4}{120}+\ldots \infty\Biggl) + \ldots\infty\ \Biggl]\)

\(=\displaystyle -\Biggl(\frac{{\theta}^2}{6}-\frac{{\theta}^4}{120}+\ldots \infty\Biggl) \Biggl[1+\frac{{\theta}^2}{12}-\frac{{\theta}^4}{240}+\ldots \ldots\infty\ \Biggl]\)

\(\displaystyle = \Biggl(-\dfrac{{\theta}^2}{6}-\dfrac{{\theta}^4}{72}+ \ldots \infty\Biggl)\)

Therefore, the identity \((*)\) can be expressed as,

\(\displaystyle \implies \Biggl(-\dfrac{{\theta}^2}{6}-\dfrac{{\theta}^4}{72}+ \ldots \infty\Biggl)= \displaystyle\Biggl(- \dfrac{{\theta}^2}{{\pi}^2}-\frac{1}{2}\dfrac{{\theta}^4}{{\pi}^4}-\ldots \infty\Biggl)+ \Biggl(-\dfrac{{\theta}^2}{2^2{\pi}^2}-\frac{1}{2}\dfrac{{\theta}^4}{2^4{\pi}^2}-\ldots\infty \Biggl)+\)

\(\displaystyle \hspace{6 cm}\Biggl(-\dfrac{{\theta}^2}{{3^2\pi}^2}-\frac{1}{2}\dfrac{{\theta}^4}{{3^4\pi}^4}-\ldots \infty\Biggl)\ldots \infty\) \(\hspace{2.5 cm}\ldots (**)\)

Again, equating the coefficients of \({\theta}^4\) from the both sides of the identity \((*)\) and \((**)\) we get

\[\dfrac{1}{2{\pi}^4}\Biggl(\dfrac{1}{1^4}+\dfrac{1}{2^4}+\dfrac{1}{3^4}+\ldots\infty\Biggl) = -\dfrac{1}{120}+\dfrac{1}{72} = \dfrac{1}{180}\]

\[ \implies\dfrac{1}{1^4}+\dfrac{1}{2^4}+\dfrac{1}{3^4}+\ldots\infty = \dfrac{{\pi}^4}{90} \]

Therefore, \(\hspace{5.5 cm}\boxed{\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{{\pi}^4}{90}}\)

So, we are done. This is all about Basel Problem. We are leaving the \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^6} = \frac{{\pi}^6}{945}\) solution for inquisitive reader. If reader is interested they can solve this.

See Also

Basel problem

On the Number of Primes Less Than a Given Magnitude

Weierstrass factorization theorem

References