Curiosity on this infinite product

[Ssnow]

TL;DR

Curiosity question on the infinite product $2\cdot 2\cdot 2\cdots$

Let us consider the infinite products $p_{n}\,=\, 2\cdot 2\cdot 2 \cdot 2 \cdots 2 \,=\, 2^n$ with $n=1,\ldots$ . Clearly $p_{n}\rightarrow +\infty$ as $n\rightarrow +\infty$. But if I put the infinity case $2\cdot 2\cdot 2 \cdot 2 \cdots \,=\, x$ I have $2\cdot x =x$ so $x=0$. It is obvious I cannot put $x=2\cdot 2\cdot 2 \cdot 2 \cdots$ and to try to seach the limit because the product diverges but has this "strange" algebraically formal result a conceptual reason to be (for example it is linked to the way to do the products ?) or it is only wrong and stop here ?
Thank you,
Ssnow

 

[FactChecker]

Summary:: Curiosity question on the infinite product $2\cdot 2\cdot 2\cdots$

Let us consider the infinite products $p_{n}\,=\, 2\cdot 2\cdot 2 \cdot 2 \cdots 2 \,=\, 2^n$ with $n=1,\ldots$ . Clearly $p_{n}\rightarrow +\infty$ as $n\rightarrow +\infty$. But if I put the infinity case $2\cdot 2\cdot 2 \cdot 2 \cdots \,=\, x$ I have $2\cdot x =x$ so $x=0$.

Or $x=\infty$. If you rule out $\infty$ here, then you are biasing the result.

 

[Ssnow]

@FactChecker thanks, sure $x=0 \vee x=\infty$. I ask for the absurd solution $x=0$ ...
Ssnow

 

[Office_Shredder]

This is a pretty classic thing where you can create fake math. The real issue is that you started off by assuming a limit exists. If there is a limit and it is L (L is a real number), then 2L=L so L=0. But this assumes the limit exists to begin with, which obviously it does not.

You can get more obvious contradictions. 1+1+1+..., If it has a limit of L then 1+L= L so 1=0.

 

[Ssnow]

@Office_Shredder I think $1+L=L$ imply that $0L=-1$ that is impossible!
In any case from something of false you can deduce everything ...
Thank you!
Ssnow