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I was wondering if the Riemann rearrangement theorem has an affect on the Euler summation transform, since, when applying it to a conditionally convergent series, the rearrangement of terms may affect the value to which the series converges to.

On the other hand, the Euler sum is supposed to, if convergent, be equal to the original series.

So... does it, or does it not affect the value it converges too?

Simply Beautiful Art
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1 Answers1

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From [1,Theorem 2.26]:

Theorem. The Euler summation method $(E,q)$ (for $q>0$) is regular. That is, $\sum_{k=0}^n a_k \xrightarrow[n\to\infty]{} s$ implies $\sum_{k=0}^n a_k \xrightarrow[n\to\infty]{(E,q)} s$.

Note that this directly implies the following:

  • If the series $\sum_n a_n$ is absolutely convergent, then its Eulerian sum (when it exists) will be the same value.

  • If the series $\sum_n a_n$ is conditionally convergent, then its Eulerian sum (when it exists) will be the same value.

and in particular:

  • Fix $\ell\in[-\infty,\infty]$. If the series $\sum_n a_n$ is conditionally convergent, there exists a rearrangement $\sum_n a_{\sigma(n)}$ converging to $\ell$, by the Riemann rearrangement theorem. Then the Eulerian sum (when it exists) of $\sum_n a_{\sigma(n)}$ will be $\ell$.

There is no contradiction: if you apply the Eulerian summation method to the conditionally convergent series $(a_n)_n$, you will get the same limit in the Eulerian sense. If you apply the Eulerian summation method to the (different) conditionally convergent series $(a_{\sigma(n)})_n$, you will get another limit than before in the Eulerian sense, because you are not considering the same series anymore in the first place.

[1] Shawyer, Bruce; Watson, Bruce (1994). Borel's Methods of Summability: Theory and Applications. Oxford University Press. ISBN 0-19-853585-6

Clement C.
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