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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?

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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 $\lim \sup s_n=\beta.$

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  • $\begingroup$ Thanks so much :)! We need $R'$ and $R''$ to be divergent in order for the last paragraph to work, right? $\endgroup$ – Abdu Magdy Jan 28 '18 at 18:53
  • $\begingroup$ We only need their positive terms (& their negative terms) to be divergent, which they are because P',P'',Q',Q'' are divergent. $\endgroup$ – DanielWainfleet Jan 29 '18 at 0:49
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For your question 1, consider the sequence $a_n=1/n$ and $b_n=-1/n$. Both series $\sum_n a_n$ and $\sum_n b_n$ diverge. However the series $\sum_n(a_n+b_n)$ converges.

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  • $\begingroup$ No, what you wrote is not correct. That difference is not defined. $\endgroup$ – Andrés E. Caicedo Jan 28 '18 at 17:07
  • $\begingroup$ Now your correction makes no sense. What do you mean by "they both diverges"? $\endgroup$ – Andrés E. Caicedo Jan 28 '18 at 17:24
  • $\begingroup$ @AndrésE.Caicedo May you suggest a correction, please? :) $\endgroup$ – Abdu Magdy Jan 28 '18 at 21:19

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