[Eulerian Mathematics] Summary of Bernoulli Series and Related Series

Bernoulli Series

When \omega=0, the first series becomes the famous Bernoulli series (also known as the Basel problem): \sum_n \frac{1}{n^2}=1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\dots Since it is related to the Bernoulli series, it is natural to start with the summation of the Bernoulli series.

The most celebrated method for summing the Bernoulli series is the "proof" given by Euler through bold conjecture and analogy. Euler considered the series: \frac{\sin \sqrt{x}}{\sqrt{x}}=1-\frac{x}{6}+\frac{x^2}{120}-\frac{x^3}{5040}+\dots From the expression on the left, it is easy to see that the roots of \frac{\sin \sqrt{x}}{\sqrt{x}}=0 are n^2 \pi^2,\,n=1,2,3,\dots.

Next, we consider a polynomial equation of degree n: 1+a_1 x+a_2 x^2 +\dots+a_n x^n=0 Assuming it has n non-zero roots x_1,x_2,\dots,x_n, we can view it as an equation for \frac{1}{x}: \left(\frac{1}{x}\right)^n+a_1 \left(\frac{1}{x}\right)^{n-1}+\dots+a_n=0 The corresponding roots become 1/x_1, 1/x_2, \dots, 1/x_n. According to Vieta’s formulas, we have: \frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_n}=-a_1

Euler’s boldness lay in directly generalizing the aforementioned Vieta’s formulas to infinite series. He argued that since the roots of \frac{\sin \sqrt{x}}{\sqrt{x}}=0 are n^2 \pi^2,\,n=1,2,3,\dots, then: \frac{1}{\pi^2}+\frac{1}{2^2 \pi^2}+\frac{1}{3^2\pi^2}+\dots=-a_1=\frac{1}{6} Thus, he told us: 1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\dots=\frac{\pi^2}{6} Bold, ingenious, and yet concise, Euler perfectly embodied the spirit of an adventurer and a captivating beauty! It is no wonder that John Stillwell, in his book Mathematics and Its History, stated that "Euler was perhaps the greatest master of infinite series."

Caution and Scrutiny

I remember when I first saw this proof, I admired Euler immensely. What kind of mind could come up with such a method! In an era when the calculus of series was just beginning to develop, he dared to generalize simple Vieta’s formulas to infinity. More importantly, he obtained the correct result! The correct result!

However, as my studies have deepened, I now look back at his process with a more critical eye—noting where it lacks rigor, where it can be generalized, and so on. Generalizing Vieta’s formulas to infinity is not impossible, but several problems need to be addressed. For instance, how can we guarantee that all roots have been listed? The roots in Vieta’s formulas can be complex, whereas n^2 \pi^2,\,n=1,2,3,\dots are at most all the real roots; are there any complex roots? If we replace \frac{\sin \sqrt{x}}{\sqrt{x}} with other functions, is it possible to obtain other similar interesting results?

Of course, I only mention this briefly here; it is not the main theme of this article, so I will not discuss it further. In fact, Euler’s method can derive rich results, and its heuristic and creative nature far outweighs its lack of rigor.

Infinite Product Results

In fact, Euler’s method tells us much more than just the special case of the Bernoulli series sum. From algebraic theory, we also know that if the n non-zero roots of the polynomial 1+a_1 x+a_2 x^2 +\dots+a_n x^n=0 are x_1,x_2,\dots,x_n, then: 1+a_1 x+a_2 x^2 +\dots+a_n x^n=a_n (x-x_1)(x-x_2)\dots (x-x_n) Substituting x=0 into this, we get: 1=a_n(-x_1)(-x_2)\dots (-x_n) Using this relationship, the above expression can be rewritten as: 1+a_1 x+a_2 x^2 +\dots+a_n x^n=\left(1-\frac{x}{x_1}\right)\left(1-\frac{x}{x_2}\right)\dots \left(1-\frac{x}{x_n}\right) Euler claimed that this could also be generalized to infinity! Therefore: \frac{\sin \sqrt{x}}{\sqrt{x}}=\left(1-\frac{x}{\pi^2}\right)\left(1-\frac{x}{2^2 \pi^2}\right)\left(1-\frac{x}{3^2\pi^2}\right)\dots Or, by changing the variable: \frac{\sin x}{x}=\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{2^2 \pi^2}\right)\left(1-\frac{x^2}{3^2\pi^2}\right)\dots Substituting x=\omega\pi, we get: \frac{\sin \omega\pi}{\omega\pi}=\left(1-\frac{\omega^2}{1^2}\right)\left(1-\frac{\omega^2}{2^2 }\right)\left(1-\frac{\omega^2}{3^2}\right)\dots This is one of the infinite products mentioned at the beginning of the article. If we replace \omega with i\omega, we get: \frac{\sin (i\omega\pi)}{i\omega\pi}=\left(1+\frac{\omega^2}{1^2}\right)\left(1+\frac{\omega^2}{2^2 }\right)\left(1+\frac{\omega^2}{3^2}\right)\dots Using the identity \sin (ix)=i\sinh x, we obtain: \frac{\sinh \omega\pi}{\omega\pi}=\left(1+\frac{\omega^2}{1^2}\right)\left(1+\frac{\omega^2}{2^2 }\right)\left(1+\frac{\omega^2}{3^2}\right)\dots This is another case of the infinite product.

Returning to Summation

Continuing our summation, if we replace \frac{\sin\sqrt{x}}{\sqrt{x}} with: \frac{\omega\sin (\pi\sqrt{x+\omega^2})}{\sqrt{x+\omega^2}\sin\omega\pi}=1+\frac{\pi\omega\cot\omega\pi-1}{2\omega^2}x+\dots This time, all the roots of the equation \frac{\omega\sin (\pi\sqrt{x+\omega^2})}{\sqrt{x+\omega^2}\sin\omega\pi}=0 are x=n^2-\omega^2,\,n=1,2,3,\dots. Therefore: \sum_n \frac{1}{n^2-\omega^2}=\frac{1-\pi\omega\cot\omega\pi}{2\omega^2} Replacing \omega with i\omega, we get: \sum_n \frac{1}{n^2+\omega^2}=\frac{1-i\pi\omega\cot i\omega\pi}{2(i\omega)^2}=\frac{\pi\omega\coth \omega\pi-1}{2\omega^2} This solves the two summation problems.

Euler, Truly a Divine Figure!

Encyclopedias mention that Euler began publishing papers at the age of 19 and continued until he was 76. He wrote a total of 886 books and papers, with over 700 papers published during his lifetime. The St. Petersburg Academy of Sciences spent a full 47 years organizing his works! It must be said that he is truly a supreme scholar worthy of the name!