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In my real analysis class, we learned how to show that an absolutely convergent series, namely $\sum\limits_{n=1}^{\infty}{a_n}$, has convergent subsequences $\sum\limits_{n=1}^{\infty}{a_n^+}$ and $\sum\limits_{n=1}^{\infty}{a_n^-}$.

Now, I'm being asked to show that if $\sum\limits_{n=1}^{\infty}{a_n}$ is conditionally convergent, then it's subsequences $\sum\limits_{n=1}^{\infty}{a_n^+}$ and $\sum\limits_{n=1}^{\infty}{a_n^-}$ diverge.

I take this to mean that each of these has infinitely many non-zero terms.

How do should I go about showing that there are in fact infinitely many non-zero terms in each subsequence causing them to diverge?

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    $\begingroup$ It takes more than infinitely many nonzero terms for a series to diverge. What about $\sum_{n\geq 1}\frac{1}{n^2}=\frac{\pi^2}{6}$? $\endgroup$ – Julien Apr 23 '13 at 23:28
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    $\begingroup$ Try to assume that if one of them converges, then both converges. And if both converges you get absolute convergence. $\endgroup$ – M.B. Apr 23 '13 at 23:33
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Simply having infinitely many non-zero terms is insufficient. There are many cases of infinite series with infinitely many non-zero terms that converge, e.g. $\displaystyle \sum_{n = 1}^\infty \frac{1}{n^p}$ for $p > 1$.

When you say that $\displaystyle \sum_{n = 1}^\infty a_n$ converges conditionally, I presume you mean that it converges but it does not converge absolutely. (As an aside, this already means that we have infinitely many non-zero terms - if there were only finitely many, it would converge absolutely, and thus be an absolutely convergent series).

Let's suppose that $\displaystyle \sum_{n= 1}^\infty a_n^-$ converges for a moment, so that we can say that

$$\sum_{n= 1}^\infty a_n^- = -L$$

for some number $-L$. We know that $\displaystyle \sum a_n$ diverges, so we must have that $\displaystyle \sum a_n^+$ diverges. But if $\displaystyle \sum a_n^+$ diverges, then $\displaystyle \sum a_n = \sum a_n^+ + \sum a_n^-$ will look like $\displaystyle \left( \sum a_n^+\right) - L$, and will still diverge. (Note that I'm being a bit abusive and leaving out details like that these are partial sums; you cannot change the order of elements in a conditionally convergent sum and expect nothing to change).

This contradicts our initial condition that $\displaystyle \sum a_n$ converges. So we must actually have that $\displaystyle \sum a_n^-$ was divergent. $\diamondsuit$

To boil down the essence of the proof, an absolutely convergent series means that all the terms get sufficiently small sufficiently fast to converge. Having a series be merely conditionally convergent means that the positive part and the negative parts cancel out a lot, and without that cancellation you get divergence. By 'cancel out a lot,' I really mean infinite cancellation since this sort of work does not worry about finite contribution. So if either the positive or negative contribution is finite, there isn't enough cancellation.

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Conditional convergence means $\sum_{n \geq 1} a_n$ converges but $\sum_{n \geq 1} |a_n| = \infty$. What you want easily follows.

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    $\begingroup$ Please keep answers constructive. $\endgroup$ – vadim123 Apr 24 '13 at 0:50
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    $\begingroup$ (+1). I guess this is the source of confusion for the question raiser assuming he/she means $a_n^{+} = max(0, a_n)$ etc. $\endgroup$ – hot_queen Apr 24 '13 at 1:21
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Let $\sum a_k=M$. Suppose to the contrary one of $\sum a_k^+$, $\sum a_k^-$ converges.

WLOG, $\sum a_k^+=N$. Then $$\sum a_k^-=\sum a_k^+-\sum a_k=N-M<\infty$$

This implies $$\sum |a_k|=\sum a_k^++\sum a_k^-=2N-M<\infty$$ which is a contradiction.

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