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

Possible Duplicate:

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?

## marked as duplicate by Zev ChonolesDec 12 '11 at 19:33

• 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
• @yoyo, what do you mean by the extension of $\sum z^n$? – David Souther Dec 12 '11 at 19:52
• @percusse: This series is convergent in the $2$-adic metric and indeed converges to $-1$. – Zhen Lin Dec 13 '11 at 2:54