The sum of an alternative series There is a series which most people will be very familiar with.
$$S=1-1+1-1+1-1+\cdots$$
We all know that it is divergent hence $S=\lim\limits_{n\to\infty}S_n=\lim\limits_{n\to\infty}\sum\limits_{i=0}^{n}(-1)^n$ has no limit value.
But on the other hand there is a method to make $S$ has a value. Looks like that:
$$\begin{equation}S+S=(1-1+1-1+\cdots)+1-1+1-1+\cdots\\
=1+(1-1)+(1-1)+(1-1)+(1-1)+\cdots\\=1\end{equation}$$
Now we have $S=\frac{1}{2}$.  
I am wondering that
Does the result imply that a divergent series can also have an exact sum?
Does the sum of the alternative series contradict the divergence?
 A: A summation method may be understood as a function
$$ \Sigma :  \mathbb{R}^{\mathbb{N}} \to \mathbb{R} \cup \{\texttt{undefined}\}. $$
For instance, the ordinary summation method $\Sigma^{\text{ord}}$ is defined as
$$ \Sigma^{\text{ord}} ((a_k)_{k=1}^{\infty}) = \begin{cases}
\lim_{n\to\infty} \sum_{k=1}^{n} a_k, & \text{if this limit converges}, \\
\texttt{undefined}, & \text{otherwise}.
\end{cases} $$
This function has two additional properties:


*

*Linearity. Suppose that $\mathbf{a} = (a_k)_{k=1}^{\infty}$ and $\mathbf{b} = (a_k)_{k=1}^{\infty}$ are summable, i.e. both $\Sigma^{\text{ord}}(\mathbf{a})$ and $\Sigma^{\text{ord}}(\mathbf{b})$ exist in $\mathbb{R}$. Then we have
$$\Sigma^{\text{ord}}(\alpha \mathbf{a} + \beta\mathbf{b}) = \alpha \Sigma^{\text{ord}}(\mathbf{a}) + \beta \Sigma^{\text{ord}}(\mathbf{b})$$
for all $\alpha, \beta \in \mathbb{R}$.

*Stability. Whenever $\mathbf{a} = (a_k)_{k=1}^{\infty}$ is summable and $\mathbf{a}' = (a_{k+1})_{k=1}^{\infty}$, then $\mathbf{a}'$ is also summable and
$$ \Sigma^{\text{ord}}(\mathbf{a}) = a_1 + \Sigma^{\text{ord}}(\mathbf{a}'). $$
Now we have two observations:


*

*$\Sigma^{\text{ord}}(1, -1, 1, -1, \cdots) = \texttt{undefined}$.

*Assume that $\Sigma$ is a summation method which satisfies both linearity and stability and that $S = \Sigma(1, -1, 1, -1, \cdots)$ is defined. Then the only possible choice of the value of $S$ is $\frac{1}{2}$, since
\begin{align*}
S
= \Sigma(1, -1, 1, -1, \cdots)
&= 1 + \Sigma(-1, 1, -1, 1, \cdots) \tag{stability} \\
&= 1 - \Sigma(1, -1, 1, -1, \cdots) \tag{linearity} \\
&= 1 - S
\end{align*}
and solving this gives $S = \frac{1}{2}$. This is essentially what you computed.
These two facts do not contradict each other.
A: For these kind of not convergent infinite series we can assign values by Cesàro sum, defined as the limit, as n tends to infinity, of the sequence of arithmetic means of the first n partial sums of the series.
In this sense we can assign to the series (aka Grandi's series) the value $\frac 12$.
