analysisconditional-convergencereal-analysissequences-and-series
Questions:
1- Why does the divergence of either one $(\sum p_n, \sum q_n)$ and the convergence of the other imply the divergence of $\sum a_n$ while the divergence of both of them doesn't imply the divergence of $\sum a_n$?
2- Why did Rudin require $\beta_1 \gt0 $, and does that imply that $\beta \gt 0$?
3- Why doesn't exist a number $\lt \alpha$ or $\gt \beta$ that can be a subsequential limit of $(25)$?
May someone explain please?
This is a theorem with a direct proof by construction.
The statement simply says "there exists a rearrangement". So the book assembles one. And it does not mean there cannot exist any other construction! The proof states one case, namely expression (25) and shows it works. That's it.
The structure of the proof is basically the following.
First, we put up Expression (25). Then we show that all its constituents indeed exist and make sense, namely $p_n, q_n, P_{m_k}, Q_{m_k}, \beta_n, \alpha_n$.
A picture is worth a thousand words.
Curves in blue and in green are sequences $\beta_n$ and $\alpha_n$, they converge to $\beta$ and $\alpha$ respectively.
Curve in red is our Expression (25). The way we painted this picture shows a few things about Expression (25).
The first terms of it are positive, so the reason why $\beta_1 >0$, and the curve first goes up. Again, in our construction we first put Expression (25) as a basis, starting point of the proof. The first terms of it are positive, so $\beta_1$ must be positive. We tweak betas and other components to make them tie in with Expression (25).
Next, $x_n > \beta_n$ and $y_n < \alpha_n$, so the red curve of Expression (25) sticks out a bit "outside" sequences $\beta_n$ and $\alpha_n$. By one last positive term and negative term $P_{m_n}$ and $Q_{k_n}$.
Differences between Expression (25) and these sequences (marked X) become smaller because $P_n \to 0$ and $Q_n \to 0$. It is precisely for this reason it becomes finally "clear" that Expression (25) cannot converge to any number greater than $\beta$ or smaller than $\alpha$.
Can we have an alternative construction? Why not!
Just change Expression (25), and tweak all other elements accordingly, e.g. make Expression (25) "expand" inside the "space" between $\beta$ and $\alpha$. Make the first terms negative numbers, the curve would first go down in this case. The differences (marked Y) would also gradually become smaller. And the whole thing would work as well.
You might ask why the theorem is stated only for the real numbers? That's because all inequalities do not make sense for the complex numbers, which do not have '<' defined.
First, let me make sure it's clear what's happening behind all those formulas. The rearrangement that allegedly works goes like this: First take just enough positive terms from your given series to produce a partial sum $>\beta$. (You can do that because the series of all the positive terms diverges.) After that, put just enough negative terms to bring the partial sum down below $\alpha$ (possible because the series of all the negative terms diverges). Then resume putting just enough positive terms to bring the partial sum back up above $\beta$. Continue working back and forth like this.
Notice that I said just enough terms at every stage. That ensures that, when you get a partial sum $s$ above $\beta$, it won't be too far above $\beta$; the difference $s-\beta$ will be at most the last term you added, because otherwise you could have stopped adding positive terms sooner. Similarly, when the partial sum goes below $\alpha$, the difference will be (in absolute value) at most the (absolute value of the) last term you added.
But your original series converged (conditionally), so the terms approach zero. That means that the amounts by which you overshoot $\beta$ and undershoot $\alpha$ are eventually arbitrarily small as you perform more and more stages of the process. And that's what those last two lines in Rudin's proof are saying is "clear".
Best Answer
Here is another way to prove Theorem 3.54.
Let $P=(p_n)_n$ be the sequence of non-negative terms of $(a_n)_n$ and let $Q=(q_n)_n$ be the sequence of negative terms of $(a_n)_n.$ Since $\sum_np_n=\infty$ and $\sum_nq_n=-\infty$ we can partition $P$ into sub-sequences $P'=(p'_n)_n,\; P''=(p''_n)_n$ and partition $Q$ into $Q'=(q'_n)_n,\; Q'' =(q''_n)_n$ such that none of $P',P'',Q',Q''$ has a finite sum.
Then merge $P',Q'$ into a single series $R'$ and merge $P'',Q''$ into a series $R''.$ ( Note that neither $R'$ nor $R''$ is required to be summable.)
$(\bullet).$ Prove that if $\lim_{n\to \infty}x_n\to 0 $ and if neither the series of non-negative terms nor the series of negative terms of $(x_n)_n$ has a finite sum then $(x_n)_n$ can be re-arranged to a series $(y_n)_n$ that sums to any given member of $[-\infty,\infty].$
Apply $(\bullet)$ to $R'$ and $R''.$ Re-arrange $R'$ to a series $S'$ converging to $\alpha.$ Re-arrange $R''$ to a series $S''$ converging to $\beta.$ Merge $S'$ and $S''$ into a single series $S.$ Then $S=(s_n)_n$ is a re-arrangement of $(a_n)_n$ with $\lim \inf s_n=\alpha$ and $\lim \sup s_n=\beta.$