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Let $\sum_{n=1}^\infty a_n$ be conditionally convergent.

Let $k_n:= \max(a_n,0),l_n:=-\min(a_n,0)$ for $n\in \mathbb{N}$ and show that

$\sum_{n=1}^\infty k_n =\infty $ and $\sum_{n=1}^\infty l_n=\infty$.

I found this exercise on some german math forum and the corresponding post was closed for the lack of answers.

Now I tried myself to it and I can visualize that the limits of those series' is $\infty$ because, since it's not absolutely convergent it is $>0$ in case of $k_n$ and also in case of $l_n$ because $\min(a_n,0)<0$ and the $(-)$ would make it positive again.

But how can I prove that mathematically? Or is the argumentation that I gave enough?

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Since $a_n = k_n - l_n$ and $|a_n| = k_n + l_n$, the condition on the series $\sum\limits_{n = 1}^\infty a_n$ imply $\sum\limits_{n = 1}^\infty (k_n - l_n)$ converges and $\sum\limits_{n = 1}^\infty (k_n + l_n)$ diverges. Since the sum and difference of a convergent series and divergent series is divergent, it follows that $\sum\limits_{n = 1}^\infty 2k_n$ is divergent and $\sum\limits_{n = 1}^\infty 2l_n$ is divergent. So both $\sum\limits_{n = 1}^\infty k_n$ and $\sum\limits_{n = 1}^\infty l_n$ are divergent.

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  • $\begingroup$ I think you mean $\sum(k_n-l_n)$ converges and $\sum(k_n+l_n)$ diverges. $\endgroup$ – user84413 Jun 21 '15 at 17:34
  • $\begingroup$ @user84413 yes that's what I meant, thank you. :) $\endgroup$ – kobe Jun 21 '15 at 17:37
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First of all we have $$\sum k_n+\sum l_n\stackrel{(1)}=\sum(k_n+l_n)\stackrel{(2)}=\sum |a_n|\stackrel{(3)}=\infty $$ where $(1)$ is valid because all $k_n,l_n$ are nonnegative, $(2)$ is valid because $k_n+l_n=|a_n|$ by definition of $k_n,l_n$, and $(3)$ is valid by assumption. Therefore, at least one of the series on the left diverges to $\infty$. If the other were finite (i.e., either $\sum k_n=K<\infty$ or $\sum l_n=L<\infty$), we'd have $$\sum a_n=\sum k_n-\sum l_n=\begin{cases}K-\infty=-\infty\text{ or}\\\infty-L=+\infty\end{cases} $$ contrary to assumption.

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  • $\begingroup$ Can I really assume that any of the two's limit of series is $\infty$? Isn't that what I want to prove? $\endgroup$ – Konstantin Jun 21 '15 at 17:30
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If $\sum a_n$ is conditionally convergent, then $\sum a_n$ converges, but $\sum |a_n|=\infty$ or, in other words

$$\lim_{n\to\infty}\sum_{m=0}^na_m=\infty$$

Let $P=\{n\in\Bbb N: a_n\ge 0\}$.

$$k_n=\begin{cases}a_n, & n\in P\\0, & n\notin P\end{cases}$$

This gives $$\sum_{n=0}^\infty k_n=\sum_{n\in P} |a_n|=\infty.$$

The last equality holds because $\lim_{n\to\infty}|a_n|=\infty$ implies that any subsequence of partial sums also approaches infinity.

The argument for $\sum l_n$ is similar.

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  • $\begingroup$ "Suppose $a_n<0$ for $n$ odd ..." - this is not justified $\endgroup$ – Hagen von Eitzen Jun 21 '15 at 17:25
  • $\begingroup$ Yes, you're right. I was assuming the series was alternating. I'll fix the argument. $\endgroup$ – Tim Raczkowski Jun 21 '15 at 17:34

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