# Series of even and odd subsequences converge

I know the result that if ##\lim a_{2n} = L = \lim a_{2n-1}##, then ##\lim a_n = L##. I'm wondering, can this be generalized to series? i.e., if ##\displaystyle \sum_{n=1}^{\infty}a_{2n-1}## and ##\displaystyle\sum_{n=1}^{\infty}a_{2n}## converge, then ##\displaystyle \sum_{n=1}^{\infty}a_{n}## converges also? I think I can see how I would prove it using the definition of convergence, but is there a slicker way to prove it using just the algebraic properties of series?

FactChecker
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If the individual series are absolutely convergent, then they can be rearranged at will and the combined series will converge. Otherwise, the convergence depends on the order of the terms and anything can happen.

If the individual series are absolutely convergent, then they can be rearranged at will and the combined series will converge. Otherwise, the convergence depends on the order of the terms and anything can happen.
So this hold only if the the two series are absolutely convergent?

fresh_42
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So this hold only if the the two series are absolutely convergent?
If they do, you can use the algebraic property you've asked for, because the order is irrelevant and we can add the infinite sums.

Otherwise you need the limit argument:

##(L_{2n}+L_{2n-1})-\varepsilon =L_{2n}-\varepsilon/2 + L_{2n-1}-\varepsilon/2 \leq S_n = S_{2n}+S_{2n-1} \leq L_{2n}+\varepsilon/2 + L_{2n-1}+\varepsilon/2 = (L_{2n}+L_{2n-1})+\varepsilon##

with the partial sums ##S_{*}## and its limits ##L_{*}##. Then ##\lim_{n \to \infty}S_n= L_{2n}+L_{2n-1}##.

WWGD
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A series can be seen as a sequence of partial sums.

FactChecker
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So this hold only if the the two series are absolutely convergent?
If any one of them is NOT absolutely convergent, then either the sum of the negative tems is ##-\infty## or the sum of the positive terms is ##+\infty##, or both. That gives enough terms of one sign to swing the partial sums as far as you want in that direction, over and over again. So that situation will also be true for the combined series. The result depends on how the terms are ordered and you can never count on the partial sums settling down. In general, that makes it impossible to apply algebraic properties.

If both series are absolutely convergent, then there is a limited ability of the terms of one sign to swing the sum in that direction. That also remains true of the combined series. The partial sums are well behaved and the final limits are the same regardless of the order of the terms. Algebraic properties can be applied.

fresh_42
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If any one of them is NOT absolutely convergent, then either the sum of the negative terms is ##\infty## or the sum of the positive terms is ##\infty##, or both.
Can you prove this, please? The minus signs don't have to be distributed in an alternating way.

FactChecker
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Can you prove this, please? The minus signs don't have to be distributed in an alternating way.
I assume that we are talking about real numbers, not complex. Then, sure, I can prove it. If ##\sum \mid a_n \mid = \infty##, then either the sum of the absolute values of the negative terms is infinite or the sum of the positive terms is infinite or both.

fresh_42
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How? If both partial series were finite, then their sum is. But how can we conclude, that their sum also has to be absolute finite?

FactChecker
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How? If both partial series were finite, then their sum is. But how can we conclude, that their sum also has to be absolute finite?
That is the opposite of what I said.

fresh_42
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That is the opposite of what I said.
No, it's the negation, not the opposite. ##\sum a_{2n} < \infty \wedge \sum a_{2n-1}<\infty \Longrightarrow \sum a_n<\infty##. You stated the stronger statement that ##\sum |a_n|<\infty## which I do not see. This is equivalent to what you said, namely ##\sum |a_n| = \infty \Longrightarrow \sum a_{2n} = \infty \vee \sum a_{2n-1}=\infty##.

I have no counterexample at hand, so I currently cannot rule out, that it is correct, but I doubt it. In any case, it is not obvious. We cannot conclude from convergence to absolute convergence, so your stronger statement needs to be proven. The point is, that convergence + not absolute convergence does not imply, that the distribution of signs goes along ##2n / 2n-1##, which is why I cannot see how you can conclude especially on these two subsequences.

For short: I see that absolute convergence is sufficient. But you claim that it is also necessary, which I do not believe.

FactChecker
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For short: I see that absolute convergence is sufficient. But you claim that it is also necessary, which I do not believe.
Oh. I see your point. I meant to say that absolute convergence is the only way it is safe to apply simple rules of algebra as the OP asked. Otherwise, one must consider the order and it gets messy.

fresh_42
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That was what I said in post #4.

mathwonk
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just off the top of my head,
how about as odd series: 1 + 1/2 - 1/3 + 1/4 - 1/5 + 1/6 - 1/7 + 1/8 -+...., and as even series: 1 - 1/4 + 1/3 - 1/8 + 1/5 - 1/12 + 1/7 - 1/16 +-......

Then the combined series seems to diverge, but both series seem to satisfy the criterion for conditional convergence. (OOPs! claim retracted below.)

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fresh_42
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just off the top of my head,
how about as odd series: 1 + 1/2 - 1/3 + 1/4 - 1/5 + 1/6 - 1/7 + 1/8 -+...., and as even series: 1 - 1/4 + 1/3 - 1/8 + 1/5 - 1/12 + 1/7 - 1/16 +-......

Then the combined series seems to diverge, but both series seem to satisfy the criterion for conditional convergence.
But if both subseries converge, then by transition to partial sums we can add those and the sum again converges. Was my argument wrong in post #4?

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