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If $\sum_{k=1}^{\infty}x_k$ is conditionally convergent, prove that $\sum_{k=1}^{\infty}x_k^+ = \infty = \sum_{k=1}^{\infty}x_k^-$ where $x_k^+$ are the positive terms and $x_k^-$ are the absolute values of negative terms of $x_k$.

Proof: Suppose $x_k$ is only conditionally summable, this means that the sequence $s_n = \sum\limits_{k=1}^{n}|x_k|$ is not convergent. Note that $s_n$ is a monotone increasing sequence that is not convergent, hence it must be unbounded. But $\sigma_n^+ = \sum\limits_{k=1}^{n}x_k^+$ and $\sigma_n^- = \sum\limits_{k=1}^{n}x_k^-$ are all subsequences of $s_n$, and they are monotone increasing as well.

Thus, they must me also unbounded and $\sum\limits_{k=1}^{\infty}x_k^+ = \infty = \sum\limits_{k=1}^{\infty}x_k^-$. $\blacksquare$

EDIT: Here's my another attempt at proving this:

Proof: Note that $s_n = \sum_{k=1}^{n}|x_k|$ is divergent to $\infty$. But $s_n' = \sum_{k=1}^{n}x_k$ converges to some finite number $L\in\mathbb{R}$. Let $\sigma_n^+ = \sum\limits_{k=1}^{n}x_k^+$ and $\sigma_n^- = \sum\limits_{k=1}^{n}x_k^-$. Then, $s_n = \sigma_n^+ + \sigma_n^- $ and $s_n' = \sigma_n^+ - \sigma_n^-$. Then, $s_n+s_n' = 2\sigma_n^+$ must diverge to infinity as $n\to\infty$ because $s_n$ diverges to $\infty$ and $s_n'$ converge to $L$. Any finite number added to infinity is also infinity. This implies that $2\sigma_n^+$ must diverge as well as $\sigma_n^+$.

Likewise, $s_n'-s_n = -2\sigma_n^-$ must diverge to $-\infty.$ And this implies $-2\sigma_n^-$ diverge to $-\infty$ and $\sigma_n^-$ diverges to $\infty$. $\blacksquare$

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    $\begingroup$ Are you taking the absolute value of the negative terms? Otherwise, $\sum x_k^- = - \infty$. $\endgroup$ – Alexis Olson Mar 8 '17 at 19:47
  • $\begingroup$ Yes, I forgot to mention that $x_k^-$ has to be absolute value of the negative terms of $x_k.$ $\endgroup$ – user3000482 Mar 8 '17 at 19:48
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    $\begingroup$ $\sigma_n^+$ and $\sigma_n^-$ are not subsequences of $s_n$. But your consideration on those partial sums is indeed a good beginning. As a possible angle of attack, you may utilize the relation $$s_n = \sigma_n^+ + \sigma_n^- \qquad \text{and} \qquad \sum_{k=1}^{n} x_k = \sigma_n^+ - \sigma_n^-$$ to reach the conclusion. $\endgroup$ – Sangchul Lee Mar 8 '17 at 19:49
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    $\begingroup$ $\sigma_n^{\pm}$ are obtained by summing certain portion of summands of $s_n$. In other words, they may be understood as partial sums for certain subsequences of $|x_k|$'s. This is not the same as picking certain terms in the sequence $(s_n)$. Let me see if I can devise a simple example that easily demonstrates this difference. $\endgroup$ – Sangchul Lee Mar 8 '17 at 19:56
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    $\begingroup$ Shorter argument: since the sequence is not absolutely summable, one of the two has to diverge; if only one diverged then the whole thing wouldn't be convergent. $\endgroup$ – Ian Mar 8 '17 at 20:51
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Notice $x_k^{+} = x_k$ if $x_k \geq 0$, and $x_k^{+}=0$ otherwise. Notice $x_k^{-} = -x_k$ if $x_k \leq 0$, and $x_k^{-}=0$ otherwise.

Put $$s_n = \sum_{k=1}^{n} |x_k|, \quad \sigma_n = \sum_{k=1}^{n} x_k, \quad \sigma_n^{+} = \sum_{k=1}^{n} x_k^{+}, \quad \sigma_n^{-} = \sum_{k=1}^{n} x_k^{-}. $$ Note $$\sigma_n^{+} = \frac{1}{2}s_n + \frac{1}{2}\sigma_n.$$ Since $s_n$ diverges and $\sigma_n$ converges, we must have that $\sigma_n^{+}$ diverges. Moreover, as $\sigma_n^{+}$ is increasing, $\sigma_n^{+} \rightarrow \infty$.

Likewise, $\sigma_n^{-} = \frac{1}{2}s_n - \frac{1}{2}\sigma_n$ leads to $\sigma_n^{-} \rightarrow \infty$.

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