# If $a_n > 0$, $\sum a_n$ diverges, and let $b_n$ such that $\frac{b_n}{a_n} \to L$ then, $\frac{\sum b_n}{\sum a_n} \to L$

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)$$$$M \geq \frac{|b_n|}{a_n} = \frac{| S^b_n - S^b_{n-1}|}{S^a_n - S^a_{n-1}} = | \frac{ S^b_n }{S^a_n } -\frac{ S^b_{n-1} }{S^a_n } |$$$$ 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.

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