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I am having trouble proving the statement below for sequences $a_n\in \mathbb R,b_n\in \mathbb R^+ $. Assuming the following limit exists then :

$$x=\lim_{n\to\infty}(\frac {a_n}{b_n}) \ne 0 \implies (\sum_{n=1}^\infty a_n =S_1 \iff \sum_{n=1}^\infty b_n =S_2) $$

Whereas $S_1,S_2 $ should simply imply that each of the series converge.

I started out by assuming that $x\ne 0$ and $\sum_{n=1}^\infty a_n$ converges. Now what came to my mind was, since the series converges it implies that $\lim_{n\to\infty}(a_n) = 0$ . Wouldn't that lead to a contradiction at this point though, since I assumed that $x\ne 0$ and if $a_n$ converges to $0$ then it must follow that $x=0$ ,right? Any hints?

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  • $\begingroup$ What do you mean with $=S_1$ and $=S_2$? $\endgroup$ – GLay Dec 2 '16 at 14:52
  • $\begingroup$ @GLay as I said, it should just mean that each of the series converge to some value. $\endgroup$ – AxiomaticApproach Dec 2 '16 at 14:53
  • $\begingroup$ And I suppose that the limit can't converge to $\infty$ $\endgroup$ – GLay Dec 2 '16 at 14:53
  • $\begingroup$ @GLay exactly... $\endgroup$ – AxiomaticApproach Dec 2 '16 at 14:54
  • $\begingroup$ Have you tried Wikipedia? en.m.wikipedia.org/wiki/Limit_comparison_test $\endgroup$ – GLay Dec 2 '16 at 14:58
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There is $N \in \mathbb N$ such that

$\frac{1}{2}|x| \le \frac{|a_n|}{|b_n|} \le \frac{3}{2}|x|$ for $n>N$. Thus

$\frac{1}{2}|x||b_n| \le |a_n| \le \frac{3}{2}|x||b_n|$ for $n>N$.

It follows:

$\sum_{n \ge 1}a_n$ converges iff $\sum_{n \ge 1}b_n$ converges

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Hint

Take $\epsilon=\frac{x}{2}$

  • use the definition of the limit.

  • use comparison test for positive series.

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