# Curiosity on this infinite product

• B
Gold Member
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##. 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

## Answers and Replies

FactChecker
Gold Member
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.

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

Office_Shredder
Staff Emeritus
Gold Member
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.

Gold Member
@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