# Show that a conditionally convergent sequence has divergent sign subsequences.

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?

• It takes more than infinitely many nonzero terms for a series to diverge. What about $\sum_{n\geq 1}\frac{1}{n^2}=\frac{\pi^2}{6}$? – Julien Apr 23 '13 at 23:28
• Try to assume that if one of them converges, then both converges. And if both converges you get absolute convergence. – M.B. Apr 23 '13 at 23:33

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.

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

• Please keep answers constructive. – vadim123 Apr 24 '13 at 0:50
• (+1). I guess this is the source of confusion for the question raiser assuming he/she means $a_n^{+} = max(0, a_n)$ etc. – hot_queen Apr 24 '13 at 1:21

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.