Can some one explain the following sentence regarding summation of divergent series?

$\text{Sum of a divergent series depends on the way how one performs the summation}.$

What does mean it?

Can someone explain it by giving an example?


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    $\begingroup$ You might like to read this: en.wikipedia.org/wiki/Riemann_series_theorem $\endgroup$ – trancelocation Feb 22 '19 at 11:22
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    $\begingroup$ The series $$ 1 - 1 + 1 - 1 +\ldots $$ comes to my mind. You can see it as $(1-1) + (1+1) + \dots$ or $1 - (1-1) - (1-1) \ldots$ $\endgroup$ – Matti P. Feb 22 '19 at 11:22

Try looking at $$\sum_{n=0}^\infty (-1)^n;$$ depending on how one groups the terms, one has $0$ or $1$:

$$ 1 + (-1 + 1) + (-1 + 1) \to 1\\ (1 - 1) + (1 -1) \cdots \to 0 .$$

  • $\begingroup$ can you explain it in terms of language ? Because you took one convergent series. But I said for divergent series $\endgroup$ – Why Feb 22 '19 at 11:26
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    $\begingroup$ With infinite series that diverge like above we can rearrange the terms in groupings that yield two values - each grouping is equally valid, but equally meaningless to the value of the sum - since it is not a convergent sum.Please take a look at the link by @trancelocation . $\endgroup$ – Chinny84 Feb 22 '19 at 11:34
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    $\begingroup$ @M.A.SARKAR If by "divergent" series you mean a sum of positive terms that does not have a finite sum, then the statement you quote is not true. Please read the Riemann series theorem Wikipedia article for details. $\endgroup$ – Somos Feb 22 '19 at 13:27

I believe it is possible that the theorem you are alluding to concerns, not divergent series, but rather conditionally convergent series.

If you insist on 'divergent', meaning 'non-convergent', then there are different ways a series may fail to converge; for some the statement given holds, for others not. Here the comment by @Somos is pertinent.


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