I believe my proposition is correct, as all of the pairs of series I've tried obey the claim. The strategy for proving this is straightforward. I assume $\sum\limits_{n=0}^{\infty} a_n$ absolutely converges and $\sum\limits_{n=0}^{\infty} b_n$ conditionally converges.

We know $\sum\limits_{n=0}^{\infty} (a_n + b_n)$ converges as both series are assumed to be convergent. The strategy is to show $\sum\limits_{n=0}^{\infty} |a_n + b_n|$ diverges. It's clear the fact $\sum\limits_{n=0}^{\infty} |b_n|$ diverges will be used somewhere, and perhaps we could prove a contradiction by assuming $\sum\limits_{n=0}^{\infty} |a_n + b_n|$ converges, we would derive the conclusion that $\sum\limits_{n=0}^{\infty} |b_n|$ converges.

I've been toying with the triangle inequality for a while but I haven't been able to get anywhere. There's also the possibility the claim is false.

  • 1
    $\begingroup$ Let $c_n=a_n+b_n$. Suppose $\sum a_n$ converges absolutely, $\sum b_n$ converges conditionally, and $\sum c_n$ converges absolutely. In other words, $\sum c_n$ and $\sum a_n$ converge absolutely, while $\sum (c_n-b_n)$ does not converge absolutely. Hmm. Do you know that the sum or difference of two absolutely convergent series is absolutely convergent? $\endgroup$ – bof Oct 25 '13 at 6:15

By contradiction, if $\sum |a_n+b_n|$ converges, since $|b_n|-|a_n|\leq |a_n+b_n|$

we have that $\sum |b_n|\leq \sum |a_n| + \sum |a_n+b_n|$

By comparison test, $\sum |b_n|$ converges.



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