# Infinite products of 2's and divisions by 2's converging to $\sqrt{2}$?

I was just doing a problem when a weird idea occurred to me. The problem required me to think about an infinite multiplications and divisions by 2:

$\dfrac{2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2 ...}{2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2 ...}$

Which I instinctively assumed was 1. However, I quickly realized this was probably wrong, since it led to some contradictions within the problem. The "crazy" idea that occurred to me is that perhaps this converges to $\sqrt{2}$ under some sort of Cesaro convergence for products, which makes total sense within the context of the problem. This is because we can express it as follows:

$\huge\prod_{n=0}^{\infty}2^{-1^{n}}=2^1\cdot2^{-1}\cdot2^1\cdot 2^{-1}\cdot2^1\cdot 2^{-1}...=2^{1-1+1-1+1-1+1-1+...}$

Where we can recognize Grandi's sum which is cesaro convergent to $\dfrac{1}{2}$, and thus:

$\huge\prod_{n=0}^{\infty}2^{-1^{n}}\text{ 'Cesaro' converges to } 2^{\frac{1}{2}}=\sqrt{2}\text{ ?}$

Note by Saúl Huerta
10 months ago

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This is meaningless, like $\displaystyle \sum_{n=1}^\infty n =-\dfrac{1}{12}$. It has to do with Riemann's $\zeta$ function.

- 9 months, 3 weeks ago

That's a really interesting application of Cesaro's convergence Saúl. Whenever infinity is involved, math can be very surprising. In this case however, Cesaro's convergence is only an attempt to solve a problem using tools which were designed for other problems. This means that the answer of $\frac{1}{2}$ was only calculated by treating the infinite sum 1 - 1 + 1 - 1 + ... like a geometric sum. However, that sum is not a true geometric sum. So while we can solve the sum as if it was a geometric sum, we won't get the real answer. And in this case, there is no answer! The series does not converge, but instead bounces back and forth between two values without ever stopping at just one. Check out this wiki for more information about the Grandi series.

- 8 months ago

The Grandi Series is a way of assigning a divergent series a finite sum, thus the above explanation or rather apparent proof is subject to the flaws that if infinty is an even entity then the product converges to 1 else if it isn't then the series converges to 2.

- 9 months, 3 weeks ago

But then how can we assign it a finite product? As I stated, for my purposes I find that $\sqrt{2}$ works well.

- 9 months, 3 weeks ago

you can assign it a finite product in the sense of cesaro convergence, however this only works when you want to analytically expand the domain of a function. However if the question was pertaining to real analysis then the above infinite product is undefined.

- 9 months, 3 weeks ago