By 苏剑林 | March 20, 2016
Recently, while calculating path integrals, I frequently encountered the following two types of infinite series:
\[\sum_n \frac{1}{n^2\pm\omega^2}\quad \text{and} \quad \prod_n \left(1\pm\frac{\omega^2}{n^2}\right)\]
Of course, one can straightforwardly compute the results using Mathematica, but I still want to know why, or at least have a general idea.
Bernoulli Series
When $\omega=0$, the first series becomes the famous Bernoulli series (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 admired 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 extending the aforementioned Vieta's formulas to infinite series. He reasoned 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 concise—Euler perfectly embodied the spirit of an adventurer and a breathtaking beauty! No wonder John Stillwell, in his "Mathematics and Its History," called Euler "perhaps the greatest scholar of series operations."
Caution and Inspection
I remember the first time I saw this proof; I admired Euler to no end. What kind of mind could come up with such a method! In an era when series operations were just beginning to develop, he dared to generalize simple Vieta's formulas to infinity. What's more, he obtained the correct result! The correct result!
However, with deeper study now, looking back at his process invites a more critical eye. For instance, where is it non-rigorous? Where can it be generalized? Extending Vieta's formulas to infinity is not impossible, but several issues must be resolved. For example, how can we guarantee that all roots have been listed? The roots in Vieta's formulas can be complex, while $n^2 \pi^2,\,n=1,2,3,\dots$ are at most all the real roots; are there complex roots? If we use a function other than $\frac{\sin \sqrt{x}}{\sqrt{x}}$, is it possible to obtain other similar interesting results?
Of course, I only mention this briefly; it is not the main theme of this article and thus won't be discussed much. In fact, Euler's method can derive yielding rich results; its inspiration and creativity far outweigh its lack of rigor.
Infinite Product Results
In fact, what Euler's method tells us is far 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 $n$-th degree equation $1+a_1 x+a_2 x^2 +\dots+a_n x^n=0$ 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 $x=0$, we get:
\[1=a_n(-x_1)(-x_2)\dots (-x_n)\]
Using this relationship, we can rewrite the equation 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 then claimed this could be extended to infinity! Thus:
\[\frac{\sin \sqrt{x}}{\sqrt{x}}=\left(1-\frac{x}{\pi^2}\right)\left(1-\frac{x}{2^2 \pi^2}\right)\left(1-\frac{x}{3^2\pi^2}\right)\dots\]
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 $\sin (ix)=i\sinh x$, 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 for the infinite product.
Back 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 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 resolves both summation problems.
Euler, Truly a God-like Man!
Encyclopedias mention that Euler began publishing papers at age 19 until age 76, writing 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 just to organize his works! It must be said, he is truly a worthy "God of Learning"!