# Is $\sum_{n=0}^{\infty}2^n$ equal to $-1$? Why? [duplicate]

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Divisibility with sums going to infinity

From Wikipedia and Minute Physics i see that the sum would be -1. I find this challenging to understand, how does a sum of positive integers end up being negative?

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## marked as duplicate by J. M., Arturo Magidin, mixedmath♦, Douglas S. Stones, Sivaram Ambikasaran Oct 19 '11 at 5:22

It's true in the $2$-adic integers. – Arturo Magidin Oct 19 '11 at 4:54
See this as well. – J. M. Oct 19 '11 at 4:54
Say you found a norm on $\mathbb{Z}$ where |2|<1. Then $\sum_{i=0}^{\infty}2^n = \frac{1}{1 - 2} = -1.$ Such a norm exists, it's called the $2$-adic norm. In fact, it is the unique norm on $\mathbb{Z}$ such that $|2| < 1.$ – jspecter Oct 19 '11 at 4:55
I miss the points about the 2-adic norms and analytic continuations on the other question! – Vinko Vrsalovic Oct 19 '11 at 5:58
Related : Divergent series and p-adics – MJD Aug 31 '14 at 21:58

It only does so formally, not literally according to the usual definition of infinite series convergence over $\mathbb{R}$ or $\mathbb{C}$. The idea is to take the series to mean the familiar geometric series and analytically continue it as a function to arguments where the original definition does not work. I attempted an answer in another question as to what analytic continuation entails exactly. One may also use a variety of summability methods to evaluate these sums, whereby we alter or redefine the partial sums in some way in order to get a finite answer. The motivation for these perhaps seemingly artificial practices is that they are examples of regularization and renormalization of expressions naturally occuring in modern physics containing divergent or infinite factors.
Suggestively, it does actually hold in a literal sense over $2$-adic integers. The fact that the series converges 2-adically follows from $|2^n|_2=2^{-n}\to0$, and if we wanted to have fun we could pull an Euleresque stunt and call the sum $S$, multiply by $2$ and say $2S=S-1\implies S=-1$.
It's all about the principle of analytic continuation. The function $f(x)=\sum_{n=0}^\infty z^n$ defines an analytic function in the unit disk, equal to the meromorphic function $g(z)=1/(1-z)$. Note that the equality $f\equiv g$ holds only in the disk, where $f$ converges absolutely. Despite this, if we naively wanted to assign any value to $f(x)$ outside of $\vert z \vert < 1$, an intuitive choice would be to set $f(2)=g(2)= -1$. Moreover, the theory of complex functions implies that this is somehow the only possible choice (for this function, at least).