Grandi's series

By Andrea Del Bravo March 6, 2022

Preface

One of my hobbies has always been math. Since I was a child I have always been fascinated by numbers: my mother told me that I did nothing but count, well before I went to primary school. In the end I really don't know why I became an engineer. This passion continues into old age, and I would like to share some results that I am proud to have discovered, although long after who first discovered it. Furthermore, I discovered that what I am about to write can be found on dozens of websites. Nonetheless, my desire to share them still remains, because they are somewhat amazing.

What is Grandi’s series?

The name is related to Guido Grandi ((monk and mathematician, professor at the University of Pisa) who first proposed it to the scientific community at the beginning of 18^th century, and can be expressed by the following formula

$G = \sum_{n = 0}^{\infty} {(- 1)}^{n}$ 1)

That is an infinitive sum of 1 and -1. The questions are: does this series have a finite result? And in case: what is the result?

Observing the series we note that if we stop at an even element the sum is 1, while if we stop at an odd element the result is 0: thus the partial sum is 1 or 0 depending on the number of elements chosen. But we have to repeat the operation for ever, and who knows what happens at infinite?

According to Cauchy’s test of convergence (see for example”Problems in Mathematical Analysis” by Boris Demidovich, I always have this book on my nightstand), the series is not converging to any number. Nevertheless some considerations can lead to another result.

For example, let’s sum to S in expressione 1), another instance of S and let’s add also a 0, which of course does not affect the result. Making clear what has been said we can write:

2G = G + 0 + G = 1 – 1 + 1 - 1 + 1 – 1 + 1 …………………………... +

0 + 1 – 1 + 1 - 1 + 1 – 1 ……………………………

Summing up each term of the first row to the second row it is now clear that the result is

2G = 1 and thus $G = \frac{1}{2}$ 2)

and it is really amazing!!!! even if formally this sum does not exist

in support of 2) there is also the Cesaro summation. Taking from Wikipedia: “In mathematical analysis, Cesàro summation (also known as the Cesàro mean) assigns values to some infinite sums that are not necessarily convergent in the usual sense. The Cesàro sum is defined as the limit, as n tends to infinity, of the sequence of arithmetic means of the first n partial sums of the series.”

Let’s make some partial sums from 1)

$G_{1} = 1$

$G_{2} = 0$

$G_{3} = 1$

$G_{4} = 0$

……….

It is clear that the partial sums are equal to 1 or 0, and thus making the means of this partial sums we obtain:

$m_{1} = \frac{G_{1}}{1} = 1$

$m_{2} = \frac{G_{1} + G_{2}}{2} = \frac{1}{2}$

$m_{3} = \frac{G_{1} + G_{2} + G_{3}}{3} = \frac{1}{3}$

$m_{4} = \frac{G_{1} + G_{2} + G_{3} + G_{4}}{4} = \frac{2}{4} = \frac{1}{2}$

$m_{4} = \frac{G_{1} + G_{2} + G_{3} + G_{4} + G_{5}}{5} = \frac{3}{5}$

…………

It is easy to see that

$\lim_{n = 1}^{\infty} m_{n} = \frac{1}{2}$ 3)

Confirming the result found in 2).

There is another clue confirming the 2). Consider the function:

$y = \frac{1}{(1 + x)}$ 4)

Let’s develop function 4) in series of functions using Taylor formula with initial point 0.

It’s easy to find that:

$y = \frac{1}{(1 + x)} = 1 - x + x^{2} - x^{3} + x^{4} - x^{5}$ …………. 5)

If in 5) we consider x = 1 we have:

$\frac{1}{2} = 1 - 1 + 1 - 1 + 1 - 1 + 1$ …….. which is the Grandi’s series. In this case the easy objection is related to interval of convergence which exclude x = 1.

Why Grandi’s series?

Good question! First answer because it is a really funny result, but, more important, we will use it to find another much more amazing result exploring the Riemann zeta function, with many confirmation coming from physics.