What is sum of reciprocals?

Image by Pete Linforth from Pixabay

In 1735, the renowned mathematician Leonhard Euler published the paper “De summis serierum reciprocarum” (On the sums of series of reciprocal) shown in Fig. 1 (click on this link for the full article). In this paper, the great master found the general closed formula for the sum:

Euler’s astonishingly clever method “has fascinated mathematicians ever since.” Euler had previously proved the Basel problem in 1734. This result extends the Basel problem from exponent 2 to any even exponent.

The three first examples are:

Note that the first case in Eq. 2 is the Basel problem, also solved by Euler (in 1734), explored in one of my recent articles.

But before undertaking Euler’s wonderful proof, the concept of Bernoulli numbers must be explained.

Bernoulli Numbers

This is the first part of the proof. We start by reminding the reader of the concept of Taylor series. Taylor series can be quickly defined as “a series expansion of a function about a point.” The subject of Taylor series is vast, so instead of exploring it here in detail, I will restrict myself to one case only, the expansion of the exponential function eˣ. It is given by:

The radius of convergence R of eˣ is R=∞. This implies that Eq. 3 is a power series that converges for all values of x ∈ ℝ.

Now, to obtain the Bernoulli numbers we perform two steps. The first is trivial: we subtract 1 from both sides of Eq. 3 and divide by x. We get:

valid for x ≠ 0. The Bernoulli numbers are defined by inverting this series as follows:

To find the Bs, we will use a mathematical trick. Since Eq. 4 and Eq. 5 are the inverse of each other (by construction), their product is 1. We can then multiply the right-hand sides of both equations and then multiply the resulting product by n!.

After some simple algebra, we obtain a nice expression that allows us the determine the Bernoulli numbers at once:

The resulting first Bernoulli numbers are then:

The Tangent Function and Its Power Series

We now provide the second part of the proof. In this section, we will need to express the tangent of x in terms of the Bernoulli numbers. Let us first consider the following identity:

which gives:

We now perform two simple steps. Substitute x on the right-hand of Eq.8 by 2ix (where i is the imaginary unit) to get:

Now, making the same substitution on the left-hand side of Eq. 8 we get:

Then we use the trigonometric identity

and Eq. 9, to arrive at:

The Cotangent Function and Its Partial Fractions

Now to the third and last part of the proof. Using partial fractions, Euler arrived at the following expansion:

We now compare Eq. 9 and Eq. 11 replacing x by πx in the former. After some simple manipulations, we arrive at this marvelous expression:

Interestingly, the is no similar formula for odd exponents (for k=1, the sum of the reciprocals of cubes is equal to a number ~1.20 called Apéry’s constant, but there is no general formula like Eq. 12). Maybe someone reading this will find that out!

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