The sum of an alternative series

Question

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?

Answer 1

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 $\frac 12$.

Answer 2

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:

1. 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 $\alpha, \beta \in \mathbb{R}$.

2. 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.