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Suppose that $\sum_{j=0}^\infty a_j$ converges and $\sum_{j=0}^\infty a_j^2<\infty$, where $a_j\in\mathbb R$, $j\ge0$. I would like to prove that $$ \biggl[\sum_{j=0}^\infty a_j\biggr]^2 =\sum_{j=0}^\infty a_j^2+2\sum_{k=1}^\infty\sum_{j=0}^\infty a_ja_{j+k}. $$

Using continuity and squaring the sum, we obtain $$ \biggl[\sum_{j=0}^\infty a_j\biggr]^2=\lim_{n\to\infty}\biggl[\sum_{j=0}^na_j\biggr]^2=\sum_{j=0}^\infty a_j^2+2\lim_{n\to\infty}\sum_{j=0}^{n-1}\sum_{k=1}^{n-j}a_ja_{j+k}. $$

The proof would be complete if I could justify the following two equalities:

  1. $\lim_{n\to\infty}\sum_{j=0}^{n-1}\sum_{k=1}^{n-j}a_ja_{j+k}=\sum_{j=0}^\infty\sum_{k=1}^\infty a_ja_{j+k}$;
  2. $\sum_{j=0}^\infty\sum_{k=1}^\infty a_ja_{j+k}=\sum_{k=1}^\infty\sum_{j=0}^\infty a_ja_{j+k}$.

Are these equalities true? How can I justify them if they're true? Or should the proof be completely different?

Any help is much appreciated!

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  • $\begingroup$ This reeks of Cesaro summability $\endgroup$ – JPi May 29 '14 at 12:56
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  1. This is justified by showing the limit: $2\lim_{n\to\infty}\sum_{j=0}^{n-1}\sum_{k=1}^{n-j}a_ja_{j+k}$ exists. It exists because the other terms in your equation are finite (i.e. their limits exist).

  2. What if you begin with the sums the desired way around. Then there's no need to swap them at the end. i.e. Instead of:

$\qquad\qquad 2\lim_{n\to\infty}\sum_{j=0}^{n-1}\sum_{k=1}^{n-j}a_ja_{j+k}$

$\qquad$ have:

$\qquad\qquad2\lim_{n\to\infty}\sum_{k=1}^{n}\sum_{j=0}^{n-k}a_ja_{j+k}$

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  • 1
    $\begingroup$ Thank you for your answer (+1). I somehow didn't notice that there's no need to change the order of infinite sums. However, I'm still not sure about the first part of the question. I understand that $\lim_{n\to\infty}\sum_{k=1}^n\sum_{j=0}^{n-k}a_ja_{j+k}$ exists, but is this limit equal to $\sum_{k=1}^\infty\sum_{j=0}^\infty a_ja_{j+k}$? Could you please explain in detail? $\endgroup$ – Cm7F7Bb Jun 4 '14 at 9:02
  • $\begingroup$ If $\sum{a_j}$ converges absolutely (i.e. all $a_j\geq0$) then there's no real issue: $a_ja_{j+k}\geq0$ and whatever order the terms of $\sum{\sum{a_ja_{j+k}}}$ are summed the same result is guaranteed. But is $\lim_{n\to\infty}\sum_{k=1}^{n}\sum_{j=0}^{n-k}a_ja_{j+k}$ the same as $\lim_{n\to\infty}\lim_{m\to\infty}\sum_{k=1}^{n}\sum_{j=0}^{m}a_ja_{j+k}$? Not necessarily if there is not absolute convergence as I understand it. Does the question say all $a_j\geq0$? $\endgroup$ – Mick A Jun 4 '14 at 15:09
  • $\begingroup$ $\{a_j\}$ are not necessarily non-negative ($a_j\in\mathbb R$, $j\ge0$). The series $\sum_{j=0}^\infty a_j$ might only converge conditionally. Probably the proof should be different... $\endgroup$ – Cm7F7Bb Jun 5 '14 at 6:50

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