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Big O and little o, absolute convergence of series where $a_n = O(b_n) $ or $a_n = o(b_n)$

Prove If $\sum_n b_n$ converges absolutely then $\sum_n a_n$ converges absolutely

BWO Contrapositive (BigO):

assume $\sum_n \left|a_n\right|$ doesn't converge and $a_n = O(b_n)$

Find $N \in \mathbb{N}$ and $M > 0$ s.t. $\forall n \ge N$, $\left|a_n\right| \le M\left|b_n\right|$ then $\displaystyle \frac{\left|a_n\right|}{M} \le \left|b_n\right|$ and since $\sum_n \left|a_n\right|$ then $\sum_n \frac{\left|a_n\right|}{M} = \frac{1}{m} \sum_n \left|a_n\right| $ does not converge and $\frac{1}{M}\sum_n \left|a_n\right| \le \sum_n \left|b_n\right| $ doesn't converge

For little o

Let $a_n = o(b_n)$ and assume $\sum_n \left|b_n\right|$ converges,

let $\epsilon >0$ find $N \in \mathbb{N}$ s.t. $n \ge N$, $\left|a_n\right| \le \epsilon\left|b_n\right| \implies \displaystyle \frac{\left|a_n\right|}{\left|b_n\right|} \le \epsilon$ hence the fraction is bounded and $\left|b_n\right| \to 0$ from the assm. above. I think I can conclude from here that $lim_{n \to \infty} \frac{\left|a_n\right|}{\left|b_n\right|} = c \le \epsilon, 0<c<\infty$ so $\sum_n\left|a_n\right|$ convergences be the limit comparison test

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  • $\begingroup$ I am not sure why you do this twice. If it is a little o it is a big O. Thus doing the first suffices. Personally I would not prove the first part by contradiction but it works. $\endgroup$ – quid Nov 14 '15 at 22:04
  • $\begingroup$ Why do we have $\lim_n \frac{|a_n|}{|b_n|} = c$ for some $ 0<c<\epsilon$ given that $a_n = o(b_n)$? Are you trying to argue by contradiction? $\endgroup$ – Xiao Nov 14 '15 at 23:10
  • $\begingroup$ @Xiao because I'm trying to use the Limit Comparison Test to achieve the desired result $\endgroup$ – oliverjones Nov 14 '15 at 23:18
  • $\begingroup$ @oliverjones then what is your definition for $a_n = o(b_n)$? Doesn't this imply $\lim_n \frac{a_n}{b_n} = 0$? given $b_n$ are positive $\endgroup$ – Xiao Nov 14 '15 at 23:27
  • $\begingroup$ @Xiao $a_n = o(b_n)$ if for every $\epsilon > 0$ there exists $N \in \mathbb{N}$ s.t. for $n \ge N$, $\left| a_n \right| \le \epsilon \left|b_n\right|$ I was trying to do the Big O proof my contrapostive. and Direct for little o $\endgroup$ – oliverjones Nov 14 '15 at 23:30

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