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This question already has an answer here:

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?

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marked as duplicate by Parcly Taxel, Lord Shark the Unknown, Professor Vector, MathOverview, Ethan Bolker Feb 24 '18 at 14:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ en.wikipedia.org/wiki/Riemann_series_theorem - i belive that answer your Q $\endgroup$ – ned grekerzberg Feb 24 '18 at 5:57
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    $\begingroup$ I'm fairly certain that you cannot reorder terms in a divergent series and have a meaningful result. $\endgroup$ – Neev Parikh Feb 24 '18 at 6:05
  • $\begingroup$ @NeevParikh. Right. $\endgroup$ – DanielWainfleet Feb 24 '18 at 9:43
  • $\begingroup$ Since as you have said yourself, $S$ has no value, it is nonsensical to write "$S+S=...$" since that is the sum of things that don't exist. $\endgroup$ – DanielWainfleet Feb 24 '18 at 9:46
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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.

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  • $\begingroup$ Nice answer. (+1). However, while I see that avoiding infinite sums is convenient for this kind of thing, I am always annoyed when people say things like "$\sum_{i=1}^{\infty}2^i$ does not exist" (or "is undefined"). There seems to be information lost by classifying a sum with value $\infty$ in the same category as one that truly does not exist due to oscillations/summation-order-dependence issues. $\endgroup$ – Michael Feb 24 '18 at 10:35
  • $\begingroup$ So I believe you could extend to $\sum:\mathbb{R}^{\mathbb{N}} \rightarrow\mathbb{R} \cup \{\infty\}\cup\{-\infty\}\cup\{\mbox{undefined}\}$, with standard arithmetic-with-infinity (i.e., for all $a \in \mathbb{R}$ we have $\infty+\infty=\infty + a = \infty$ and so on) with caveats that the linearity property only is required to hold when it avoids cases of $\infty -\infty$ or $0\cdot \infty$. $\endgroup$ – Michael Feb 24 '18 at 10:47
  • $\begingroup$ I deliberately did not include $\pm\infty$ because the listed two properties of $\Sigma$ only concerns algebraic structure and not the topological structure (whereas $\pm\infty$ are topological in nature). For instance, any summability method which assigns a finite value $S$ to $(1,2,4,8,\cdots)$ must give $S=-1$ and this may be the preferred choice in the theory of consideration. $\endgroup$ – Sangchul Lee Feb 24 '18 at 11:29
  • $\begingroup$ Nice observation that $$S(\{2^i\}_{i=0}^{\infty}) = 2^0 + 2S(\{2^i\}_{i=0}^{\infty}) \implies S \in \{\infty, -\infty, -1, \mbox{undefined}\}$$ (and taking the finite value gives $-1$). Is there a cute way to evaluate $S(\{i\}_{i=1}^{\infty})$, assuming it is assigned a finite value? $\endgroup$ – Michael Feb 24 '18 at 21:48
  • $\begingroup$ Assume that $\Sigma$ is a summability method satisfying both linearity and stability and that $(1,2,3,\cdots)$ is $\Sigma$-summable (i.e., $\Sigma(1,2,3,\cdots) \in \mathbb{R}$). Then the stability tells that $(2,3,4,\cdots)$ is $\Sigma$-summable and by the linearity $(1,1,1,\cdots) = (2,3,4,\cdots)-(1,2,3,\cdots)$ must be $\Sigma$-summable. But this implies $$\Sigma(1,1,1,\cdots) = 1 + \Sigma(1,1,1,\cdots), $$ which is impossible. The lesson is that we need to give up either linearity or stability to assign a finite value to $1+2+3+\cdots$, and stability is often discarded. $\endgroup$ – Sangchul Lee Feb 25 '18 at 4:14
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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 $\frac 12$.

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