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Let $∑a_n $ be a series of real numbers which converges, but not absolutely.

Let $ p_n = \frac{a_n +|a_n|}{2} \ge 0, \ q_n = \frac{a_n - |a_n|}{2} \ \le 0 $.

Let $ P_n = \sum_{k=1}^{n}{p_k}, \ Q_n = \sum_{k=1}^{n}{q_k} $.

Show that ${ \lim_{n\to\infty} \frac{Q_n}{P_n} } = -1$.

I was able to show that both $ {Q_n} $ and $ {P_n} $ diverges. But I couldn't proof the ratio between them (I'm not sure if it's helpful). How can it be proofed?

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Note that $$P_n + Q_n = \sum_{k=1}^n a_k$$ so $$ 1+ \frac{Q_n}{P_n} = \frac{\sum_{k=1}^n a_k}{P_n} \rightarrow \frac{const.}{\infty} = 0 $$

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    $\begingroup$ Thanks. but does the fact that we add "1" to the ratio of the sequences change our original problem? I mean we prove that 1 + qn/pn goes to 0, but we know nothing about qn/pn... $\endgroup$
    – Nave Tseva
    May 8 '19 at 10:10
  • $\begingroup$ We just subtract 1 from both sides. The limit of a sum is a sum of the limits, in this case $$ 0 = \lim_{n\rightarrow\infty}(1+\frac{Q_n}{P_n}) = (\lim_{n\rightarrow\infty}1)+(\lim_{n\rightarrow\infty}\frac{Q_n}{P_n}) = 1+ \lim_{n\rightarrow\infty}\frac{Q_n}{P_n}$$ $\endgroup$ May 8 '19 at 10:32
  • $\begingroup$ Thanks! It might be out of this question's scope, but how do we show that $P_n$ necessarily goes to infinity? Is it enough to say the the series $P_n$ is monotonically increasing and diverges? $\endgroup$
    – Nave Tseva
    May 8 '19 at 13:43
  • $\begingroup$ Yes that is enough. If it were monotonically increasing and bounded, then it would be convergent. $\endgroup$ May 8 '19 at 14:30

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