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If a series converges absolutely, then any sub-series also converges absolutely.

1) is it true? 2) is there a simple "one line" argument to prove it?

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    $\begingroup$ Partial sums (of the absolute values) are bounded non-decreasing. $\endgroup$ Mar 28, 2015 at 15:28

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If the “subseries” has general term $(a_{n_k})$, then for the $k$-th partial sum of the subseries we have $$ \sum_{i=0}^k|a_{n_i}|\le\sum_{i=0}^{n_k}|a_i|\le\sum_{i=0}^{\infty}|a_i| $$

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Hint. The answer is yes, since for any sub series $\sum u_{\phi(n)}$, you may write $$ \sum\left| u_{\phi(n)}\right|\leq \sum\left| u_{n}\right|. $$

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  • $\begingroup$ This is unclear to me; what are the bounds of the sums? If they're infinite, this is circular, and if they're finite, it would seem to hinge on $u_n$ being decreasing, which is not necessarily true. $\endgroup$ Mar 28, 2015 at 20:26
  • $\begingroup$ @Meelo Let's take it as finite sums. If the RHS converges, then the LHS converges (the partial sum is increasing and bounded): this gives an answer the question in the title. $\endgroup$ Mar 28, 2015 at 20:55

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