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We all know there exists absolutely convergent series: it is one where it does not matter what the order of the series is upon summation of the infinite series.

I was wondering if there was some kind of similar property for when we assign values to divergent sums such that it does not matter which order we sum the terms in.

Questions

Do "absolutely divergent series" (series which whose value assigned does not matter on the ordering) exist? Is there any quick method to identify such divergent series?

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    $\begingroup$ Um... surely $1 + 2 + 3 +4 +.....$? Or $1 + 1/2 + 1/3 + 1/4 + ....$? Or $1 + 1 + 1 + 1+.....$? $\endgroup$ – fleablood Sep 6 '17 at 6:55
  • $\begingroup$ Is there a quick method to identify such a series? $\endgroup$ – drewdles Sep 6 '17 at 7:03
  • $\begingroup$ How do you mean assign values to divergent series? $\endgroup$ – kingW3 Sep 6 '17 at 9:02
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    $\begingroup$ @kingW3 .. through analytic continuation or regularization $\endgroup$ – drewdles Sep 6 '17 at 9:09
  • $\begingroup$ @AnantSaxena Consider adding this information into the question as to what you meant by assigning value. $\endgroup$ – kingW3 Sep 6 '17 at 9:58
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You're missing the point of analytic continuation. Analytic continuation of a series doesn't have anything to do with the series i.e you're not changing the order of terms or manipulating the series in any way.

What you're actually doing is finding an analytic function which agrees with the given series where it converges but is also defined somewhere where the series diverges.

I've copied the stuff below the line from Wikipedia


If $f$ is an analytic function on a non-empty open subset $U$ of $\Bbb{C}$ then analytic continuations are unique in the following sense: if $V$ is the connected domain of two analytic functions $F_1$ and $F_2$ such that $U$ is contained in $V$ and for all $z$ in $U$ $$F_1(z)=F_2(z)=f(z)$$ then $$F_1=F_2$$ on all of $V$.

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Two simple classes of examples:

  • Any sequence of positive numbers whose sum in one ordering diverges will diverge in every ordering
  • Any sequence of numbers whose limit is not zero in one ordering will not have limit zero in any ordering, and thus forms a divergent sum
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  • $\begingroup$ "ny sequence of positive numbers whose sum in one ordering diverges will diverge in every ordering", youtube.com/watch?v=w-I6XTVZXww $\endgroup$ – jimjim Sep 6 '17 at 20:36
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For series in $\mathbb{R}$ the situation is very clear: Let

$$ P = \sum_{n=1}^{\infty} a_n^{+} \quad \text{and} \quad N = \sum_{n=1}^{\infty} a_n^{-}, $$

where $a_n^+ = \max\{a_n, 0\}$ and $a_n^- = \max\{-a_n, 0\}$. Then

  1. If $P < \infty$ and $N < \infty$, the series $\sum_{n=1}^{\infty} a_n$ is absolutely convergent.

  2. If $P = \infty$ and $N < \infty$, then for any permutation $\sigma : \mathbb{N} \to \mathbb{N}$ we have $\sum_{n=1}^{\infty} a_{\sigma(n)} = \infty$.

  3. If $P < \infty$ and $N = \infty$, then for any permutation $\sigma : \mathbb{N} \to \mathbb{N}$ we have $\sum_{n=1}^{\infty} a_{\sigma(n)} = -\infty$.

  4. If $P = \infty$ and $N = \infty$, then there exist permutations $\sigma$ and $\tau$ on $\mathbb{N}$ such that both $\sum_{n=1}^{\infty} a_{\sigma(n)}$ and $\sum_{n=1}^{\infty} a_{\tau(n)}$ exist in $[-\infty, \infty]$ but they do not equal: $\sum_{n=1}^{\infty} a_{\sigma(n)} \neq \sum_{n=1}^{\infty} a_{\tau(n)}$.

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