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There is a series which most people will be very familiar with.


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?


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

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  • $\begingroup$ en.wikipedia.org/wiki/Riemann_series_theorem - i belive that answer your Q $\endgroup$ – ned grekerzberg Feb 24 '18 at 5:57
  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

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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