You are currently browsing the tag archive for the ‘Bernoulli’ tag.

As we discussed in the previous post, the Bernoulli brothers Johann and Jakob were both intrigued by the divergence of the harmonic series. Jakob in particular was fascinated by infinite series in general, and he turned his attention to a problem which had been posed by Pietro Mengoli in 1644.

Find the exact sum (and not just an estimate) of the series

$\displaystyle \begin{array}{rcl} \sum^{\infty}_{n=1}\frac{1}{n^2}= 1+ \frac{1}{4}+ \frac{1}{9}+ \ldots. \end{array}$

In 1655, John Wallis, an English mathematician,  had communicated that he had found the sum to three decimal places but he was unable to say anything more concrete. Jakob Bernoulli wrote about this problem in 1689 and it came to be known as the Basel problem, after the hometown of the Bernoullis as well as that of the eventual solver of the problem, Leonhard Euler. There’s a lot of literature available on this problem and some of what we say below is from William Dunham’s award-winning book titled  Euler: The master of us all and the paper by Raymond Ayoub titled Euler and the zeta function.
It was known that the series converges but finding the exact sum proved to be remarkably difficult. Many prominent mathematicians, including Leibnitz, Mengoli and the Bernoullis brothers had tried their hand at solving it but had failed. Indeed, after growing increasingly frustrated with his failure, Jakob said the following about this problem [Dunham]:

“If anyone finds and communicates to us that which thusfar has eluded our efforts, great will be our gratitude”

In 1721, a young Euler began studying mathematics under the mentorship of Jakob’s younger brother Johann Bernoulli (who was one of the most prominent mathematicians of the world at that time). In all probability, it was Johann who first told Euler about the Basel problem but it is unclear exactly when he did so. Nonetheless, by 1728 Euler had started working on the problem. It was around the same time when Daniel Bernoulli (Johann’s son) wrote to Christian Goldbach that he had found an approximate value of the sum of the series (the value he gave was ${8/5}$). Goldbach replied that he had found that the sum is between ${\frac{41}{35}= 1.64}$ and ${\frac{5}{3}= 1.66}$. One of Euler’s earliest attempts was to find numerical approximations of some of the partial sums of the series but these were not too helpful. Indeed, the sum of the first thousand terms is ${1.643937}$ but this is only accurate up to the first two digits (the problem being that the series converges very slowly).

We now discuss Euler’s solution to the Basel problem.

Theorem (Euler, 1734) ${\sum^{\infty}_{n=1}\frac{1}{n^2}= \frac{\pi^2}{6}}$.

This is truly a remarkable result and anyone who sees it for the first time cannot help but be amazed. Euler’s proof is a shining example of his ingenuity and mathematical prowess. In order to prove the result he used the well-known series expansion for ${\sin x}$:

$\displaystyle \begin{array}{rcl} \sin x= x- \frac{x^3}{3!}+ \frac{x^5}{5!}- \frac{x^7}{7!}+\ldots \end{array}$

Next, he considered the “infinite polynomial”

$\displaystyle \begin{array}{rcl} P(x)= 1- \frac{x^2}{3!}+ \frac{x^4}{5!}- \frac{x^6}{7!}+\ldots \end{array}$

and observed that ${P(x)= \frac{\sin x}{x}}$ for ${x \neq 0}$. Now, the roots of the polynomial are given by all ${x}$ such that ${\sin x= 0}$ (and ${x \neq 0)}$. This gives us ${x= \pm n\pi}$ and this means that for each ${n \geq 1}$, there are two roots. Using this observation, Euler factored ${P(x)}$ as follows:

$\displaystyle \begin{array}{rcl} 1 &-&\frac{x^2}{3!}+ \frac{x^4}{5!}- \frac{x^6}{7!}-\ldots= P(x)\\ &=& \left(1- \frac{x}{\pi}\right)\left(1- \frac{x}{-\pi}\right)\left(1- \frac{x}{2\pi}\right)\left(1- \frac{x}{-2\pi}\right)\ldots\\ &=&\left(1- \frac{x^2}{\pi^2}\right)\left(1- \frac{x^2}{4\pi^2}\right)\left(1- \frac{x^2}{9\pi^2}\right)\left(1- \frac{x^2}{16\pi^2}\right)\ldots \end{array}$

The next step illustrates Euler’s foresight and his genius, for he expanded the right-hand side as follows

$\displaystyle \begin{array}{rcl} 1- \frac{x^2}{3!}+ \frac{x^4}{5!}- \frac{x^6}{7!}+\ldots= 1- \left(\frac{1}{\pi^2}+ \frac{1}{4\pi^2}+ \frac{1}{9\pi^2}+ \ldots\right)x^2+ \ldots, \end{array}$

where the remaining terms in the expansion on the right are not relevant to the problem at hand. Euler then compared the coefficients of ${x^2}$ in the above equation to get

$\displaystyle \begin{array}{rcl} \frac{1}{3!}= \left(\frac{1}{\pi^2}+ \frac{1}{4\pi^2}+ \frac{1}{9\pi^2}+ \ldots\right) \end{array}$

which gave him the celebrated solution to the Basel problem. This was a triumphant moment for Euler and it truly established his reputation as the foremost mathematician at that time. While the general reaction on the result was that of amazement,  a few remained skeptical. In his proof Euler had performed manipulations on infinite series considering them as polynomials and Daniel Bernoulli objected to this approach. Euler himself was less convinced about the rigor of the method so he devised other proofs justifying the formula. He also looked at the $p$-series ${\sum^{\infty}_{n=1}\frac{1}{n^p}}$ for ${p> 2}$.

These ideas were used by Bernhard Riemann in the 19th century in studying the Riemann-zeta function

$\displaystyle \begin{array}{rcl} \zeta(s)= \sum^{\infty}_{n=1}\frac{1}{n^s}, \end{array}$

in connection with his investigation of the distribution of primes. In this notation, the solution of the Basel problem reads as ${\zeta(2)= \pi^2/6}$. Remarkably enough, in 1740 Euler gave a more general formula for calculating ${\zeta(2k)}$ if ${k\geq 1}$ is an integer:

$\displaystyle \begin{array}{rcl} \zeta(2k)= \frac{(-1)^{k-1}B_{2k}(2\pi)^{2k}}{2(2k)!}, \end{array}$

where ${B_{2k}}$ are the Bernoulli numbers.

Since then a number of modern proofs have popped up. A nice summary of some of the available proofs is given by Robin Chapman in his article titled Evaluating ${\zeta(2)}$. Interestingly, while we know the sum of the ${p}$-series when ${p}$ is positive and even, not a lot is known about the sum of the series when ${p}$ is odd. As far as I know, there is no exact formula for ${\zeta(3)}$.

In this post we look at the harmonic series

$\displaystyle \begin{array}{rcl} \sum^{\infty}_{n=1}\frac{1}{n}= 1+ \frac{1}{2}+ \frac{1}{3}+\ldots. \end{array}$

Students often have difficulty coming to grips with this fact that this series diverges when they encounter it in an elementary calculus course. The result may seem somewhat counter-intuitive at first, since as ${n}$ gets larger, only very small terms get added to the existing sum. The harmonic series serves as an example of a divergent series whose ${nth}$ term tends to zero, thus justifying the statement that the converse of the following theorem does not hold in general.

Theorem If a series ${\sum^{\infty}_{n=1} a_n}$ converges then ${\lim_{n\rightarrow \infty}a_n= 0}$.

How do we show that the harmonic series diverges? One of the standard proofs presented in many textbooks goes like this.
Let ${s_n= \sum^{n}_{k=1}\frac{1}{k}}$ be the ${n}$-th partial sum. Observe that

$\displaystyle \begin{array}{rcl} s_1 &=& 1+ 0\left(\frac{1}{2}\right)\\ s_2 &=& 1+ \frac{1}{2}= 1+ 1\left(\frac{1}{2}\right)\\ s_4 &=& 1+ \frac{1}{2}+ \left(\frac{1}{3}+ \frac{1}{4}\right)\\ & > & 1+ \frac{1}{2}+ \left(\frac{1}{4}+ \frac{1}{4}\right)= 1+ 2\left(\frac{1}{2}\right)\\ s_8 &=& 1+ \frac{1}{2}+ \left(\frac{1}{3}+ \frac{1}{4}\right)+ \left(\frac{1}{5}+ \frac{1}{6}+ \frac{1}{7}+ \frac{1}{8}\right)\\ & > & 1+ \frac{1}{2}+ \left(\frac{1}{4}+ \frac{1}{4}\right)+ \left(\frac{1}{8}+ \frac{1}{8}+ \frac{1}{8}+ \frac{1}{8}\right)= 1+ \frac{3}{2}, \end{array}$

and in general,
$\displaystyle \begin{array}{rcl} s_{2^{n}} \geq 1+ \frac{n}{2} \end{array}$

This implies that ${\lim_{n\rightarrow \infty}s_{2^{n}}= \infty}$ and hence the harmonic series diverges (why?). This proof (which was given by Oresme around 1350) is fairly simple and easy to understand by most students and it makes you wonder if there are any other proofs out there which are more “hands on” and which only use techniques that first-year calculus students can understand. It turns out that there are at least thirty-nine proofs of the divergence of the harmonic series which can be found in the excellent article by Kifowit and Stamps titled The Harmonic Series Diverges Again and Again and by Kifowit titled More Proofs of the Divergence of Harmonic Series.

A proof by Johann Bernoulli

In this section we will discuss a remarkable proof of the divergence of the harmonic series which first appeared in the work of Jakob Bernoulli who generously attributed it to his brother Johann Bernoulli. For a fascinating account of the history behind this proof and more, please look at the following article:

William Dunham, The Bernoullis and the harmonic series, College Mathematics Journal, 18 (1987), 18-23.

Before looking at Bernoulli’s proof, let’s put things in perspective a little bit. The proof was devised around 150 years before the concept of convergence of series was made precise (by Cauchy) via the notion of convergence of partial sums. It is therefore understandable that Bernoulli looked at infinite series in a somewhat “naive” way (at least compared to the modern notion of convergence of a series). Nonetheless, the proof is quite refreshing and ingenious because it is based on, strangely enough, the convergence of the series ${\sum^{\infty}_{n=1}\frac{1}{n(n+1)}}$. In what follows, we present a modern version of Bernoulli’s proof given in the aforementioned article by Kifowit and Stamps.
The series

$\displaystyle \begin{array}{rcl} \frac{1}{2}+ \frac{1}{6}+ \frac{1}{12}+ \frac{1}{20}+\ldots= \sum^{\infty}_{n=1}\frac{1}{n(n+1)}= \sum^{\infty}_{n=1}\left(\frac{1}{n}- \frac{1}{n+1}\right) \end{array}$

is a convergent telescoping series whose sum is 1. In a similar manner, it can be shown that ${\sum^{\infty}_{n=k}\frac{1}{n(n+1)}= \frac{1}{k}}$ for ${k=1, 2, 3, \ldots}$. Bernoulli’s proof is by contradiction. Suppose that the harmonic series converges and suppose that its sum is ${S}$. Then,
$\displaystyle \begin{array}{rcl} S &=& 1+ \frac{1}{2}+ \frac{1}{3}+ \frac{1}{4}+ \frac{1}{5}+ \frac{1}{6}+ \frac{1}{7}+ \frac{1}{8}+ \ldots\\ &=& 1+ \frac{1}{2}+ \frac{2}{6}+ \frac{3}{12}+ \frac{4}{20}+ \frac{5}{30}+ \frac{6}{42}+ \frac{7}{56}+ \ldots\\ &=& 1+ \left(\frac{1}{2}+ \frac{1}{6}+ \frac{1}{12}+ \frac{1}{20}+ \ldots\right)+ \left(\frac{1}{6}+ \frac{1}{12}+ \frac{1}{20}+ \frac{1}{30}+ \ldots\right)\\ &+& \left(\frac{1}{12}+ \frac{1}{20}+ \frac{1}{30}+ \frac{1}{42}+\ldots\right)+ \ldots\\ &=& 1+ 1+ \frac{1}{2}+ \frac{1}{3}+ \ldots\\ &=& 1+ S. \end{array}$

This final statement ${S= 1+ S}$ is a contradiction. Thus the harmonic series diverges.