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Let $(a_i)_{i=1}^\infty$ be a sequence of positive numbers such that $\sum_1^\infty a_i < \infty$. What can we say about the double series $\sum_{i, j=1}^\infty a_{i+ j}^p\ ?$ Can we find some values of $p$ for which it converges? I'm especially interested in $p=2$.

Intuitively I'm inclined to think that the series converges for $p \ge 2$. This intuition comes from the continuum analog $f(x)= x^a, \quad x>1$: if $a<-1$ we have

$\int_1^\infty f(x)\ dx < \infty$

and $F(x, y)=f(x+y)$ is $p$-integrable on $(1, \infty) \times (1, \infty)$ for $p \ge 2$.

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To sum up, the result is false in general (see first part of the post below), trivially false for nonincreasing sequences $(a_n)$ if $p<2$ (consider $a_n=n^{-2/p}$) and true for nonincreasing sequences $(a_n)$ if $p\ge2$ (see second part of the post below).


Rearranging terms, one sees that the double series converges if and only if the simple series $\sum\limits_{n=1}^{+\infty}na_n^p$ does. But this does not need to be the case.

To be specific, choose a positive real number $r$ and let $a_n=0$ for every $n$ not the $p$th power of an integer (see notes) and $a_{i^p}=i^{-(1+r)}$ for every positive $i$. Then $\sum\limits_{n=1}^{+\infty}a_n$ converges because $\sum\limits_{i=1}^{+\infty}i^{-(1+r)}$ does but $na_{n}^p=i^{-pr}$ for $n=i^p$ hence $\sum\limits_{n=1}^{+\infty}na_n^p$ diverges for small enough $r$.

Notes:
(1) If $p$ is not an integer, read $\lfloor i^p\rfloor$ instead of $i^p$.
(2) If the fact that some $a_n$ are zero is a problem, replace these by positive values which do not change the convergence/divergence of the series considered, for example add $2^{-n}$ to every $a_n$.


To deal with the specific case when $(a_n)$ is nonincreasing, assume without loss of generality that $a_n\le1$ for every $n$ and introduce the integer valued sequence $(t_i)$ defined by $ a_n\ge2^{-i} \iff n\le t_i. $ In other words, $ t_i=\sup\{n\mid a_n\ge2^{-i}\}. $ Then $\sum\limits_{n=1}^{+\infty}a_n\ge u$ and $\sum\limits_{n=1}^{+\infty}na_n^p\le v$ with $ u=\sum\limits_{i=0}^{+\infty}2^{-i}(t_i-t_{i-1}),\quad v=\sum\limits_{i=0}^{+\infty}2^{-ip-1}(t_{i+1}^2-t_i^2). $ Now, $u$ is finite if and only if $\sum\limits_{i=0}^{+\infty}2^{-i}t_i$ converges and $v$ is finite if and only if $\sum\limits_{i=0}^{+\infty}2^{-ip}t_i^2$ does. For every $p\ge2$, one sees that $2^{-ip}t_i^2\le(2^{-i}t_i)^2$, and $\ell^1\subset\ell^2$, hence $u$ finite implies $v$ finite.

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    Caro Didier, in questo momento sono in una summer school molto impegnativa per me. Quando $f$iniro', la settimana prossima, potro' tornare sul $f$orum e ri$f$lettere bene sulle questioni aperte. Grazie di tutto il tempo che mi hai dedicato in questa e in tante altre domande. Buona vita!2011-08-24