3
$\begingroup$

Understanding when $$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{a_0 + a_1+ \cdots + a_n}{b_0+b_1+ \cdots + b_n}.$$

That is, supposing the first limit exists, what properties must be satisfied by the sequences $\{a_n\}_{n \ge 0}, \{b_n\}_{n \ge 0}$ such that we may conclude that the above equality holds? I tried using the inequality $\frac{a_n}{b_n} + \frac{a_{n-1}}{b_{n-1}} < \frac{a_n+a_{n-1}}{b_n + b_{n-1}}$, which holds when our sequences consist of positive real numbers, having in mind the notion of applying induction but didn't get anywhere with that idea. This isn't a homework assignment, I just want to know the answer.

$\endgroup$
  • $\begingroup$ Oh cool I'll check out the Stolz-Cesaro Theorem. $\endgroup$ – Sid Jul 29 '17 at 23:21
  • $\begingroup$ Basically, let $sa_n=a_0+\dots+a_n$ and $sb_n=b_0+\dots+b_n$, then apply Stolz-Cesaro. $\endgroup$ – Simply Beautiful Art Jul 29 '17 at 23:21
  • $\begingroup$ @SimplyBeautifulArt Thanks for the help. Follow up question: Why, assuming the first limit exists, does $\lim_{n \to \infty} \frac{a_0+a_1+ \cdots + a_n}{b_0+b_1+\cdots+b_n} = \lim_{x \to 1}\frac{a_0+a_1x+a_2x^2+\cdots}{b_0+b_1x+b_2x^2+\cdots}$? It's intuitively "obvious" but I can't formulate it in terms of epsilons and deltas. $\endgroup$ – Sid Jul 30 '17 at 0:06
  • $\begingroup$ See Section 2 of this paper for my motivations google.com/… $\endgroup$ – Sid Jul 30 '17 at 0:58
  • $\begingroup$ Hm, I'm not entirely sure how to prove it. I get the feeling it follows from Abel's theorem or something, assuming that $\sum a_k$ and $\sum b_k$ exist. $\endgroup$ – Simply Beautiful Art Jul 30 '17 at 12:28