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Let $\sum_{n=1}^\infty a_n<\infty$ be a complex convergent series

is the series $\sum_{n=1}^\infty (a_n \cdot \sum_{m=n+1}^\infty a_m)$ calculable ?

thanks for any suggestion

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    what's happen in this case? – 2017-01-17
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    Are you asking if the series is convergent, or do you really want us to be able to calculate the sum? – 2017-01-17
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    any information as welcome, the calculation is the best – 2017-01-17
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    $a:=\sum\limits_{n=1}^\infty a_n$ => Your series is equal to $a^2-\sum\limits_{n=1}^\infty a_n \sum\limits_{k=1}^n a_k$ . – 2017-01-17
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    @user90369 Assuming the derivation is correct, why does the second term converge? – 2017-01-17
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    @user90369 thanks, but it is not useful for me because i started from that relation, it moves my problem – 2017-01-17
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    @MateyMath If all you have is that $\sum_{n = 1}^{\infty} a_n$ is convergent, then the other series can be divergent. – 2017-01-17
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    O.k. . Perhaps your question becomes clearer if you explain what you mean with calculation here. :-) – 2017-01-17
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    @DanielFischer yes you are right so i need as much information as possible – 2017-01-17
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    @user90369 if there's any form to exprime that by a – 2017-01-17
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    To express the series by $a$ ? No, I don't think so. Can you please clearify $\sum\limits_{n=1}^\infty a_n < \infty$ ? For complex values your inequation doesn't exist. Do you mean $|\sum\limits_{n=1}^\infty a_n | < \infty$ or $\sum\limits_{n=1}^\infty |a_n| < \infty$ ? Or something else ? – 2017-01-17
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    i mean convergent – 2017-01-17
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    Thanks! But I see *Clement C.* has answered your question. :-) – 2017-01-17

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Assuming that both the following series are convergent $$ \sum_{n\geq 1}a_n,\qquad \sum_{n\geq 1}a_n^2 $$ we have the following symmetry trick: $$ \sum_{n\geq 1}a_n\sum_{m>n}a_m = \sum_{1\leq n < m}a_n a_m = \sum_{1\leq m < n}a_n a_m = \frac{1}{2}\left[\left(\sum_{n\geq 1}a_n\right)^2-\sum_{n\geq 1}a_n^2\right]. $$

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    @MateyMath: you're welcome. – 2017-01-18
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This is not convergent in general.

  • If your series is absolutely convergent, then indeed your new series will be absolutely convergent as well, by comparison (this is straightforward).

  • Otherwise, it may not be. Consider for instance the alternating series given by the general term $a_n \stackrel{\rm def}{=} \frac{(-1)^n}{\sqrt{n}}$ for $n\geq 1$. We have that $\sum_{n=1}^\infty a_n$ converges, but $$ \sum_{k=n+1}^\infty a_k = \frac{(-1)^{n+1}}{2\sqrt{n}} + o\left(\frac{1}{\sqrt{n}}\right) $$ (if I am not mistaken); from which you get $$ \sum_{n=1}^\infty \sum_{k=n+1}^\infty a_n a_k = -\frac{1}{2} \sum_{n=1}^\infty \frac{1}{n} \xrightarrow[n\to\infty]{} -\infty $$ by comparison.

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    thanks so much @Clement C. – 2017-01-17
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    Absolute convergence is not strictly needed, we may compute such double series under weaker assumptions, for instance $\{a_n\}_{n\geq 1}\in\ell^1\cap\ell^2$. – 2017-01-17
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    True... I didn't mean to give a full characterization, just mention that extra assumptions were needed (the simplest being absolute convergence). – 2017-01-17