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Let $H$ be a Hilbert space and $\sum_k x_k$ a convergent infinite sum in it. Lets say we partition the sequence $(x_k)_k$ in a sequence of blocks of finite length and change the order of summation only in those blocks, like this (for brevity illustrated only for the first two blocks) $$(x_1,\ldots,x_k,x_{k+1},\ldots,x_{k+l},\ldots )$$ becomes

$$(x_{\pi(1)},\ldots,x_{\pi(k)},x_{\gamma(k+1)},\ldots,x_{\gamma(k+l)},\ldots ),$$ where $\pi$ and $\gamma$ are permutations.

If we denote the elements of the second sequence with $x'$, does anyone know, what will happen to the series $\sum _k x'_k$ in this case, i.e. will it converge ? If yes, to what value ?

(This question is a refinement of this question, where it was shown, that the two sums are equal, if they both converge; here we know only that one of the series converges)

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1 Answer 1

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If the sizes of the blocks are bounded, then the permuted summation also converges to the same limit.

More generally, let $\sigma : \Bbb{N}_0 \to \Bbb{N}_0$ be a permutation such that $M = \sup_{n\in\Bbb{N}_0}|\sigma(n) - n|$ is finite. You can easily verify that this generalizes the former special case. Let $x'_n = x_{\sigma(n)}$ be the permuted sequence and $A_n = \{0, \cdots, n\}$. Now, from the assertion we find that

  1. $\sigma(A_n) \subset A_{n+M}$ and likewise $\sigma^{-1}(A_n) \subset A_{n+M}$, and
  2. $A_{n+M} \setminus \sigma(A_n) \subset A_{n+M}\setminus A_{n-M}$.

The first claim is trivial. To prove the second one, just observe that for $n > M$,

$$ A_{n+M}\setminus \sigma(A_n) \subset A_{n+M}\setminus \sigma(\sigma^{-1}(A_{n-M})) = \{n-M+1, \cdots, n+M\}. $$

Now we are ready. Let $(s_n)$ and $(s'_n)$ denote the partial sum of $(x_n)$ and $(x'_n)$, respectively. Then

$$\begin{align*}|s_n - s'_n| &\leq |s_{n+M} - s_n| + |s_{n+M} - s'_n| \\ &\leq \sum_{k=n+1}^{n+M} |x_k| + \sum_{k \in A_{n+M}\setminus \sigma(A_n)}|x_k| \\ &\leq \sum_{k=n+1}^{n+M} |x_k| + \sum_{k = n-M+1}^{n+M} |x_k|. \end{align*}$$

Now there are only $3M$ terms of $(x_k)$, each of which converges to 0 as $n \to \infty$. Thus by squeezing lemma, $|s_n - s'_n| \to 0$. This shows that

$$ \sum_n x'_n = \lim_n s'_n = \lim_n s_n = \sum_n x_n.$$

If $M$ is unbounded, you can easily devise a counter-example. For example, if we take the sizes of the blocks as $2, 4, 6, \cdots$ to the conditionally convergent series

$$ 1 - 1 + \frac{1}{2} - \frac{1}{2} + \frac{1}{2} - \frac{1}{2} + \frac{1}{3} - \frac{1}{3} + \frac{1}{3} - \frac{1}{3} + \frac{1}{3} - \frac{1}{3} + \cdots = 0 $$

and permute the order of the sequence in each block as

$$ 1 - 1 + \frac{1}{2} + \frac{1}{2} - \frac{1}{2} - \frac{1}{2} + \frac{1}{3} + \frac{1}{3} + \frac{1}{3} - \frac{1}{3} - \frac{1}{3} - \frac{1}{3} + \cdots, $$

it oscillates between 0 and 1.

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