Is $\sum \limits_{n=0}^{\infty}2^n=-1$?

Asked Dec 12 '11 at 19:30

Active Dec 13 '11 at 01:02

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Possible Duplicate:
Infinity = -1 paradox

MinutePhysics has what initially looks like a divergent series summing to -1. The youtube comments are... lacking in clarity. The argument MinutePhysics loosely makes is

$1+2+4+8+16...$

$=(1)(1+2+4+8+16...)$

$=(2-1)(1+2+4+8+16...)$

$=(2+4+8+16+32...) - (1+2+4+8+16...)$

$=-1 + (2+4+8+16+32...)-(2+4+8+16...)$

Is this proof correct, and if not, what is the error?

edited Apr 13 '17 at 12:19

Community

asked Dec 12 '11 at 19:30

David Souther

3

the extension of $\sum z^n$ to the rest of the plane is $1/(1-z)$ which is $-1$ at $z=2$ – yoyo Dec 12 '11 at 19:34
2

Manipulated series must be convergent to render the arithmetic sensible. – Dec 12 '11 at 19:34
@yoyo, what do you mean by the extension of $\sum z^n$? – David Souther Dec 12 '11 at 19:52
@David: he meant "analytic continuation". You do have what's called a "geometric series"... – J. M. ain't a mathematician Dec 13 '11 at 01:01
@percusse: This series is convergent in the $2$-adic metric and indeed converges to $-1$. – Zhen Lin Dec 13 '11 at 02:54
@ZhenLin Yep, but I didn't want to complicate things. That's why I didn't write the series are not convergent. – Dec 13 '11 at 04:51
Hmmmm... Im going to have to learn the p-adic system. All I know is that the geometric series diverges, therefore playing with the arithmetic is bound to create paradoxes. – CogitoErgoCogitoSum Feb 12 '13 at 02:15

Is $\sum \limits_{n=0}^{\infty}2^n=-1$?

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