9
$\begingroup$

Let $ \sum_{n=1}^\infty a_n $ be a conditionally convergent series. Then the series $\sum_{n=1}^\infty b_n$ of positive terms of $\{a_n\}_{n=1}^\infty$ and $\sum_{n=1}^\infty c_n$ of negative terms are divergent.

My proof: if $\sum_{n=1}^\infty a_n$ is conditionally convergent, then $\sum_{n=1}^\infty |a_n| = \infty$. Let's suppose that $\sum_{n=1}^\infty b_n$ and $\sum_{n=1}^\infty c_n$ are convergent, so since they are composed of positive and negative terms respectively, then $\sum_{n=1}^\infty |b_n| < \infty$ and $\sum_{n=1}^\infty |c_n| <\infty$ and we have that $$ \sum_{n=1}^\infty |a_n| = \sum_{n=1}^\infty |b_n| + \sum_{n=1}^\infty |c_n| < \infty $$ which is absurd since we supposed that the series was conditionally convergent.

Edit

Can I define $b_n := \left\{\begin{array}{ll} a_n &\text{if } a_n > 0 \\ 0 &\text{otherwise} \end{array} \right.$ and $c_n$ analogously for negative terms and say that $\sum a_n = \sum b_n + \sum c_n $ and, since $\sum a_n$ converges then both $\sum b_n, \sum c_n$ converge? Are the new series equal to the series of positive (and negative) terms?

Is this correct? Thanks in advance

$\endgroup$
2
  • 4
    $\begingroup$ Arguing by contradiction (or proving the contrapositive), you'd start by assuming that at least one of $\sum b_n$ and $\sum c_n$ is convergent. But it's easy to show that if one is convergent, then they both are. $\endgroup$ Commented Jan 4, 2013 at 20:23
  • $\begingroup$ @V.Galerkin: Note that what you wrote in your edit would contradict what you are trying to show. Also, $\sum c_n<\infty$ doesn't mean much. $\endgroup$ Commented Jan 4, 2013 at 21:57

3 Answers 3

10
$\begingroup$

Note that $$b_n=\dfrac{a_n+|a_n|}{2} \;\;(\geqslant 0), \\ c_n=\dfrac{a_n-|a_n|}{2} \;\;(\leqslant 0) $$ and $$c_n=a_n-b_n.$$ If $\sum a_n$ converges (conditionally) and $\sum b_n$ is convergent (absolutely) then $\sum {c_n}$ is convergent (absolutely). Because $a_n=b_n+c_n$ then $\sum a_n$ must be absolutely convergent, which contradicts to its conditional convergence.

$\endgroup$
4
  • $\begingroup$ Thanks! My edit express the same idea right? Is there any difference? $\endgroup$ Commented Jan 4, 2013 at 21:07
  • $\begingroup$ @V.Galerkin It was defined in the problem that $(b_n)$ is the sequence of positive terms of the sequence. So, you should introduce new notation for the "zero padded" sequence; define say $\tilde b_n$ either as above or, equivalently, as in your edit. A similar remark holds for $(c_n)$. You might also point out that padding a series with zeroes does not alter whether it converges. $\endgroup$ Commented Jan 4, 2013 at 21:33
  • $\begingroup$ Strochyk, how did you come up with the idea $$b_n=\dfrac{a_n+|a_n|}{2} \;\;(\geqslant 0), \\ c_n=\dfrac{a_n-|a_n|}{2} \;\;(\leqslant 0)?$$ I would have never thought about doing that. $\endgroup$
    – Lays
    Commented Apr 21, 2013 at 3:28
  • $\begingroup$ When I was a student (long time ago) I had very good teachers. $\endgroup$ Commented Apr 21, 2013 at 17:30
1
$\begingroup$

As Jonas Meyer said, you haven't proved what you set out to prove, that neither the $b$ series nor the $c$ series converges. What you have proved, though, is that they can't both converge. So you could complete the proof if you could somehow exclude the possibility that one of the two converges and the other diverges.

$\endgroup$
0
$\begingroup$

This is incorrect, because the negation of [$P$ and $Q$] is [(not $P$) or (not $Q$)]. So if you are going to argue by contradiction, you would suppose that $\sum b_n$ converges OR $\sum c_n$ converges.

$\endgroup$
2
  • 1
    $\begingroup$ Also, are we allowed to speak about split series $a_n$ like the OP divided it? $\endgroup$
    – Mikasa
    Commented Jan 4, 2013 at 20:29
  • 1
    $\begingroup$ @Babak: In that context, what the OP wrote about the case where $\sum b_n$ and $\sum c_n$ both converge is valid, including the "splitting", although I do not know what level of justification is expected and what prior results are assumed. The splitting the OP wrote in the later edit, however, is incorrect, as indicated for instance by the fact that it contradicts the problem statement. (This was after your comment.) $\endgroup$ Commented Jan 4, 2013 at 21:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .