# Proof of the ratio test

by Bipolarity
Tags: proof, ratio, test
 P: 783 I am trying to understand something in the proof of the ratio test for series convergence. If $a_{n}$ is a sequence of positive numbers, and that the ratio test shows that $\lim_{n→∞}\frac{a_{n+1}}{a_{n}} = r < 1$, then the series converges. Apparently, the proof defines a number R : r0 such that $\frac{a_{n+1}}{a_{n}} < R$ for all n>N. It need not to be true in the case where n=N, right? Up to this part I get. But then it concludes from the above that, there exists a positive N such that $a_{N+1}  P: 439 Hmm, think about what it means for the limit to exist. I think that might be the missing piece that shows why it works. Because if you believe it exist, there is an n greater than or equal to N such that (a_(n+1))/(a_n) is less than or equal to R for n greater than N. Then from there you can rewrite the inequality to get what you got. P: 783  Quote by MarneMath Hmm, think about what it means for the limit to exist. I think that might be the missing piece that shows why it works. Because if you believe it exist, there is an n greater than or equal to N such that (a_(n+1))/(a_n) is less than or equal to R for n greater than N. Then from there you can rewrite the inequality to get what you got. Are you sure the part in bold is correct? Doesn't the limit definition exclusively use greater than? Because that is precisely what I don't fully understand. BiP Mentor P: 18,278 Proof of the ratio test  Quote by Bipolarity Are you sure the part in bold is correct? Doesn't the limit definition exclusively use greater than? Because that is precisely what I don't fully understand. BiP It doesn't really matter. The statement: For all [itex]\varepsilon>0$, there exists an N such that for all $n\geq N$ holds that $|a_n-a|<\varepsilon$.
For all $\varepsilon>0$, there exists an N such that for all $n> N$ holds that $|a_n-a|<\varepsilon$.