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I've been struggling with the following exercise. We're doing the consequences of Riemann's theorem for series, and I've proved that a rearrangement of a serie can diverge to $-\infty$ and $+\infty$, but I don't know how to deal with this problem.

Let $\sum a_n$ be a conditionally convergent series (that is, convergent, but not absolutely convergent). Let $a<b$ be real numbers. Prove that exists a rearrangement and two sequences $N_k$ and $M_k$ such that $$\sum_{n=1}^{N_k} a_{\sigma(n)}<a$$ and $$\sum_{n=1}^{M_k} a_{\sigma(n)}>b,$$ when $k\to \infty$.

Any hint will be appreciated. Thanks.

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  • $\begingroup$ You haven't show us the conditions on $a_n$. $\endgroup$
    – Thomas Andrews
    Oct 21, 2015 at 0:37
  • $\begingroup$ If I understand the question correctly, let $a_n$ and $b_n$ denote the positive and negative terms of the series, respectively. Assuming the series is conditionally convergent, we have $\sum a_n = \infty$ and $\sum b_n = -\infty$. So, if we form a partial sum using only $b_n$'s, we can eventually make the sum less than any given $a$. Similarly, if we form a partial sum using only $a_n$'s, we can eventually make the sum greater than any given $b$. $\endgroup$
    – user169852
    Oct 21, 2015 at 0:37
  • $\begingroup$ @ThomasAndrews Sorry, I completely forgot to put them. $\endgroup$
    – Relure
    Oct 21, 2015 at 0:37
  • $\begingroup$ @Bungo They're sequences. $\endgroup$
    – Relure
    Oct 21, 2015 at 0:39
  • $\begingroup$ @Bungo My professor used (exactly) that notation on the exercise, I can't understand the meaning of your expression too. So I think that he want to sum up the numbers from $n=1$ to $N_k$, but he wants to note that one sequence is different to the other (the $M_k$ one). $\endgroup$
    – Relure
    Oct 21, 2015 at 0:41

1 Answer 1

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Here is a sketch:

Observation: since $\sum a_n$ is conditionally convergent, it's not hard to prove that the derived series $\sum a^+_n$ and $\sum a^{^{-}}_n$ both diverge.

Now, to begin, add, in order, just enough of the first positive terms of $\sum a_n$ so that their sum is greater than $b$. This is possible by the observation. Then we will need $M_1$ terms of $\sum a_n$ to do this. Therefore, the first partial sum of our rearrangment is $S_{M_{1}}=\sum_{n=1}^{M_1} a_{\sigma(n)}>b$.

Now, to $S_1$, add just enough negative terms from the original series so that the sum is less than $a$. We will need a total of $N_1$ terms of the original series to do this. Therefore, the first partial sum here is $S_{N_{1}}=\sum_{n=1}^{N_1} a_{\sigma(n)}<a$.

Now proceed inductively: having chosen $M_k$ and $N_k$ such that $S_{M_{k}}=\sum_{n=1}^{M_k} a_{\sigma(n)}>b$ and $S_{N_{k}}=\sum_{n=1}^{N_k} a_{\sigma(n)}<a,\ $ to $S_{N_{k}}$ we add just enough positive terms from the original series, so that $S_{M_{k+1}}=\sum_{n=1}^{M_{k+1}} a_{\sigma(n)}>b$, and to this series just enough negative terms so that $S_{N_{k+1}}=\sum_{n=1}^{N_{k+1}} a_{\sigma(n)}<a$.

By construction, these series satisfy the requirements of the exercise.

The point of this is that the sequence of partial sums $S_k=\left \{ S_{M_1},S_{N_1},\cdots , S_{M_k},S_{N_k}\right \}_{k\geq 1}$ is a rearrangement of the original series such that $\limsup S_k=b$ and $\liminf S_k=a$.

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  • $\begingroup$ Lol that's a great screen name $\endgroup$
    – AnotherPerson
    Oct 21, 2015 at 2:36

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