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Shows that if $a_n > 0$, $\sum a_n$ diverges, and let $b_n$ such that $ \frac{b_n}{a_n} \to L$ then, $\frac{\sum b_n}{\sum a_n} \to L$.

My attempts: let $\sum^N_{n=1} a_n = S^a_N$, $\sum^N_{n=1} b_n = S^b_N$ and $\frac{\sum^N_{n=1} b_n}{\sum^N_{n=1} a_n} = S_N $. As $\frac{b_n}{a_n}$ is convergent, it is also limited (by $M$). And $S^a_n$ is increasing.

a)\begin{equation} M \geq \frac{|b_n|}{a_n} = \frac{| S^b_n - S^b_{n-1}|}{S^a_n - S^a_{n-1}} = | \frac{ S^b_n }{S^a_n } -\frac{ S^b_{n-1} }{S^a_n } | \end{equation} And then got stuck...

b) For $S_N$ to converge $\frac{b_n}{S^a_N}$ must go to $0$, as $S_N = \sum_{n=1}^N \frac{b_n}{\sum^N_{n=1} a_n}$. I know that $\frac{1}{S^a_N} \to 0$ I tried to find a way to use Dirichlet, but got stuck again.

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  • $\begingroup$ Have you heard of Cesàro-Stolz? en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem In your case, try to prove: $\liminf \frac{b_n}{a_n} \leq \liminf \frac{\sum{b_n}}{\sum{a_n}}\leq \limsup \frac{\sum{b_n}}{\sum{a_n}}\leq \limsup \frac{b_n}{a_n}$ $\endgroup$ – lc2r43 Apr 14 at 23:26
  • $\begingroup$ Thanks! That theorem solves my problem. $\endgroup$ – Marlon Apr 16 at 22:12

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