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I found that if the series $\sum {a_n}$ convergences conditionally, there exists the convergent rearrangement series $\sum {b_n}$.

For example, Let rearrangement of $$\sum {(-1)^{n+1} \frac{1}{n}} = ln2$$ is

$$ 1 - \frac{1}{2} - \frac{1}4 + \frac{1}3 - \frac{1}6 - \frac{1}8 + \cdots $$

Can I roughly associate some terms like this?

$$ \left( \frac{1}1 - \frac{1}2 - \frac{1}4 \right) + \left( \frac{1}3 - \frac{1}6 - \frac{1}8 \right ) + \cdots $$

so, the normal term is

$$ \frac{1}{2n - 1} - \frac{1}{4n-2} - \frac{1}{4n} $$

like this?

If I calculate like this, the result is $\frac{1}{2} ln2$

But, can I assure that the associative property is valid for this rearranged series? I even don't know this series converges or diverges. Would you please give me the right, strict and reliable answer?

I want to show the possibility of associative property in this rearranged series. Because normally, the associative property is invalid in infinite series. Is it OK to associate some terms if the original form is convergent conditionally?

S. Yoo
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  • What do you mean by "the associative property"? – TiddSchmod Mar 28 '19 at 14:51
  • If I can parenthesize some terms, I can find the normal term and the result of the series. – S. Yoo Mar 28 '19 at 14:57
  • I want to show the possibility of associative property in this rearranged series – S. Yoo Mar 28 '19 at 15:02
  • Normally, the associative property is not valid in infinite series. – S. Yoo Mar 28 '19 at 15:03
  • Actually, you can associate since the terms converge to zero; what you get when you associate as above is $S_{3n}$ which converges to your limit; then since $S_{3n+1}-S_{3n} = \frac{1}{2n - 1}$, it follows $S_{3n+1}$ converges to the same limit and same for $S_{3n+2}$; the paradoxes mentioned in the linked answer $1-1+1-1..$ are due to the fact that in those cases the general term doesn;t converge to zero and then it is legitimate that $S_{2n}=0$ converges to zero and $S_{2n+1}=1$ converges to one, but that just shows that the partial sums of the series have two limit points and nothing more – Conrad Mar 29 '19 at 03:06

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