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

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

Staff Emeritus