How to derive $\sum_{n=0}^\infty 1 = -\frac{1}{2}$ without zeta regularization On Wikipedia we find $$\displaystyle \bbox[5px,border:1px solid #F5A029]{1 + 1 + 1+\dots =\sum_{n=0}^\infty 1 = -\frac{1}{2}}$$ using (the rather complicated) zeta-function regularization.  I asking for an elementary derivation possibly based on the idea that on average
$$ \sum_{n=0}^\infty (-1)^n  = 
\begin{cases}
1 & \text{ if }n\text{ is odd} \\
0 & \text{ if }n\text{ is even} \\
\end{cases}  = \frac{1}{2} $$
There is a blog discussion that uses very general "cutoff" functions, but I am having a hard time specializing to the case at hand.

A serious question is to qualify where the basic properties of addition break down:


*

*$(a+b)+c = a+(b+c)$ associativity

*$a+b = b+a$ commutativity

*$a + 0 = a = 0 + a$ addition by zero
This example obviously shows we can't use these axioms infinitely many times without generating contradictions.  A related question  even has $\sum 1 = 0$.

For $f(x)=\sum_0^\infty a_n x^n$, a real number $R$, $R\neq 1$, and
$g(x)=f(x)-Rf(x^2)$, if $g(1)$ is Abel summable, that is,
$g(1)=\lim_{x\to 1^-} g(x)$, an elementary Ramanujan sum of $f(1)$ is
defined by 
$$
f(1)=\frac{g(1)}{1-R}.
$$
Inversely, for an Abel summable series $g(1)=\sum_0^\infty g_n$ and
a real number $R$, $R\neq 1$, the corresponding elementary Ramanujan
summable series $f(1)$ is defined by
$$
a_0=\frac{g_0}{1-R}~,~~a_{2n+1}=g_{2n+1}~,~~a_{2(n+1)}=g_{2(n+1)}+Ra_{n+1}~.
$$
$f(1)$ is an element of the linear space of elementary Ramanujan summable
series corresponding to $R$, the Ramanujan class $R$.
If $f(1)$ is an element of two distinct classes, $f(1)$ is Abel summable,
that is, an element of the class $R=0$. 
One should not naively sum series of distinct classes (here).
For $F_k(x)=f(x^k)$, $F_k(1)=f(1)$ since $\lim_{x\to 1^-} g(x^k)=g(1)$.
$F_k(1)=a_0+0+\cdots +0+a_1+0+\cdots +0+a_2+\cdots$, with $k-1$ zeros between
$a_n$ and $a_{n+1}$. $F_k(1)$ is an element of the class $R$, as $f(1)$.  
For $f'(x)=x\frac{df(x)}{dx}$, $f'(1)$ is an element of the class $2R$. 
Consider $f(x)=\frac{x}{1-x}=x+x^2+x^3+\cdots$,
$g(x)=\frac{x}{1+x}=x-x^2+x^3-\cdots$ and $R=2$.
$g(1)$ is Abel summable to 1/2 and the elementary sum of $f(1)$
is -1/2.
$f'(1)=1+2+3+\cdots = -1/12$ is an element of the class $R=4$ and
$f''(1)=1+4+9+\cdots = 0$ is an element of the class $R=8$ etc.
Consider now $f(x)=x+2x^2+4x^4+\cdots$, $g(x)=x$ and $R=2$.
The elementary sum of $f(1)=\sum_1^\infty a_n$, with $a_n=n$
if $n$ is a power of 2 and $a_n=0$ if not, is $-1$.
$f'(1)=\sum_1^\infty a_n^2=-1/3$, $f''(1)=\sum_1^\infty a_n^3=-1/7$ etc.
To verify that $1-1+2-2+3-3+\cdots=1/8$,
consider 
$$
f(x)=x-x^2+2x^3-2x^4+3x^5-3x^6+\cdots=\frac{1}{(1+x)^2}\frac{x}{1-x}~,
$$
$$
g(x)=\frac{x-x^2-x^3-x^4}{(1+x)^2(1+x^2)^2}
$$
and $R=2$.
Edit.
The consistency of (iterated) elementary Ramanujan summation with analytic continuation of Dirichlet series is shown here (this question is deleted!), here  (this question is deleted!) and here.
