How to sum this divergent series to 1/2?

Question

Let $S=1+(-1)+1+(-1)+...=\sum_{n=0}^{\infty}(-1)^n$, under which sense does it sum up to 1/2?

I think it shall be one among Euler summation, Borel summation, Cesàro summation, or a subset, but I don't really know these summation methods. Would anyone be kind enough to elaborate them for me, or just point out which summation should it be and I'll look it up in the books?

Answer 1

The answer is a Cesàro Summation. Check the example on Grandi's Series.

Answer 2

Formally, $$\frac{1}{1-(-1)}=1-1+1-1+\cdots,$$ and the left hand side is equal to $1/2$, although this is not really valid hence use of the word formal. Note that $$\frac{1}{1-x}=x^0+x^1+x^2+x^3+\cdots,$$ where $|x|<1$.

Answer 3

Consider $$ S(x)=\sum_{k=0}^\infty x^k $$ Clearly $xS(x)+1=\sum_{k=0}^\infty x^k=S(x)\implies S(x)=\frac{1}{1-x}$. However, you may note that convergence tests yield that $S$ only converges for $|x|<1$ so it is not accurate to say $S(-1)=\frac{1}{1-(-1)}$. What we can do is say $$ \lim_{x\to -1}\sum_{k=0}^\infty x^k=\frac12 $$

Answer 4

Using just the axioms of a summation "machine" that are $$S(\{\alpha a_i+\beta b_i\}_{i\in\mathbb{N}})=\alpha S(a_1,a_2,a_3,...)+\beta S(b_1,b_2,b_3,...)\\S(x_1,x_2,x_3,...)=x_1+S(x_2,x_3,...)$$ we have $S=1-1+1-1+...=1+(-1)\cdot S\Rightarrow 2S=1\Rightarrow S=\frac 1 2$

In order to be precise, before being allowed to use Euler summation, Borel summation, you have to verify that these methods fullfill the axioms of such a machine.

Reference: Carl Bender, Lecture on physical mathematics (http://www.youtube.com/watch?v=VvqeJkT3uyo)