[Euler's Mathematics] A Summary of the Bernoulli Series and Related Series

[欧拉数学]伯努利级数及相关级数的总结 苏剑林 · 2016-03-20 · Mathematics series euler mathematics

Recently, while computing path integrals, I kept running into the following two 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, Mathematica can crank out the results without much fuss, but I still wanted to know why — at least to have some rough sense of it.

The Bernoulli series

When $\omega=0$, the first series becomes the famous Bernoulli series

$$\sum_n \frac{1}{n^2}=1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\dots$$

Since this is related to the Bernoulli series, it's natural to start from the summation of the Bernoulli series itself. more

The most celebrated method for summing the Bernoulli series is surely the "proof" that Euler gave through a bold guess 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 left-hand side, it's 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, consider the degree-$n$ polynomial equation

$$1+a_1 x+a_2 x^2 +\dots+a_n x^n=0$$

Suppose it has $n$ nonzero roots $x_1,x_2,\dots,x_n$. Then we can regard it as an equation in $\frac{1}{x}$

$$\left(\frac{1}{x}\right)^n+a_1 \left(\frac{1}{x}\right)^{n-1}+\dots+a_n=0$$

with the corresponding roots becoming $1/x_1,1/x_2,\dots,1/x_n$. By Vieta's formulas, we then get

$$\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_n}=-a_1$$

What made Euler bold was that he directly extended this instance of Vieta's formulas to an infinite series. He said: since the roots of $\frac{\sin \sqrt{x}}{\sqrt{x}}=0$ are $n^2 \pi^2,\,n=1,2,3,\dots$, then we should have

$$\frac{1}{\pi^2}+\frac{1}{2^2 \pi^2}+\frac{1}{3^2\pi^2}+\dots=-a_1=\frac{1}{6}$$

and from this he arrived at

$$1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\dots=\frac{\pi^2}{6}$$

Bold, ingenious, and yet elegant — Euler perfectly embodies the spirit of the adventurer along with an intoxicating sense of beauty! Small wonder that John Stillwell, in his Mathematics and Its History, calls "Euler perhaps the greatest master of series manipulation."

Caution and scrutiny

I still remember the first time I saw this proof — I was utterly in awe of Euler. What kind of mind could come up with a method like this! At a time when the theory of series was only just getting off the ground, he dared to extend the simple Vieta's formulas to infinity. And worse still, he arrived at the correct result! The correct result!

But having studied more deeply since then, when I revisit his derivation now, I look at it with a more critical eye — where does it lack rigor, where could it be generalized, and so on. Extending Vieta's formulas to infinity is not impossible in principle, but several issues need to be resolved. For instance, how do we guarantee that we've listed all the roots? The roots in Vieta's formulas can be complex, whereas $n^2 \pi^2,\,n=1,2,3,\dots$ at most has all its real roots accounted for — are there complex roots too? And instead of $\frac{\sin \sqrt{x}}{\sqrt{x}}$, could we use some other function and obtain other similarly interesting results?

Of course, this is just a passing remark here — it's not the main theme of this post, so I won't dwell on it further. In fact, Euler's method spawns a wealth of results, and its inspirational, creative value has long since overshadowed its lack of rigor.

Results for infinite products

In fact, what Euler's method tells us goes far beyond this one special case of the Bernoulli series. From algebra we also know that if the degree-$n$ polynomial equation $1+a_1 x+a_2 x^2 +\dots+a_n x^n=0$ has $n$ nonzero roots $x_1,x_2,\dots,x_n$, 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 $x=0$ in gives

$$1=a_n(-x_1)(-x_2)\dots (-x_n)$$

Using this relation, we can rewrite the equation above 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 again says: this can be extended to infinity! So we have

$$\frac{\sin \sqrt{x}}{\sqrt{x}}=\left(1-\frac{x}{\pi^2}\right)\left(1-\frac{x}{2^2 \pi^2}\right)\left(1-\frac{x}{3^2\pi^2}\right)\dots$$

Or, changing variables

$$\frac{\sin x}{x}=\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{2^2 \pi^2}\right)\left(1-\frac{x^2}{3^2\pi^2}\right)\dots$$

and substituting into $x=\omega\pi$, 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 exactly one of the infinite products mentioned at the very beginning of this post. If we replace $\omega$ with $i\omega$, we get

$$\frac{\sin (i\omega\pi)}{i\omega\pi}=\left(1+\frac{\omega^2}{1^2}\right)\left(1+\frac{\omega^2}{2^2 }\right)\left(1+\frac{\omega^2}{3^2}\right)\dots$$

and 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 the other case of the infinite product.

Back to summation

Continuing with our summation, if we replace $\frac{\sin\sqrt{x}}{\sqrt{x}}$ with

$$\frac{\omega\sin (\pi\sqrt{x+\omega^2})}{\sqrt{x+\omega^2}\sin\omega\pi}=1+\frac{\pi\omega\cot\omega\pi-1}{2\omega^2}x+\dots$$

then 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 $x=n^2-\omega^2,\,n=1,2,3,\dots$, and so we have

$$\sum_n \frac{1}{n^2-\omega^2}=\frac{1-\pi\omega\cot\omega\pi}{2\omega^2}$$

Replacing $\omega$ with $i\omega$ gives

$$\sum_n \frac{1}{n^2+\omega^2}=\frac{1-i\pi\omega\cot i\omega\pi}{2(i\omega)^2}=\frac{\pi\omega\coth \omega\pi-1}{2\omega^2}$$

and this settles both of the summation problems.

Euler, truly a mathematical demigod!

According to the encyclopedia, Euler started publishing papers at age 19 and, right up to age 76, wrote a total of 886 books and papers, over 700 of which were published during his lifetime. It took the St. Petersburg Academy of Sciences a full 47 years to organize all of his works! One really has to say — this is what it truly means to be a "god of learning"!