Proof of the ratio test

Bipolarity

I am trying to understand something in the proof of the ratio test for series convergence.
If [itex]a_{n}[/itex] is a sequence of positive numbers, and that the ratio test shows that [itex] \lim_{n→∞}\frac{a_{n+1}}{a_{n}} = r < 1[/itex], then the series converges.

Apparently, the proof defines a number R : r<R<1, and then shows that there exists a N>0 such that [itex]\frac{a_{n+1}}{a_{n}} < R [/itex] 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
[itex] a_{N+1}<a_{n}R [/itex] which does not follow due to the statement in bold.

Could someone please point out where I am wrong so I can continue this theorem without any qualms? Thanks!

BiP

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.

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

micromass

It doesn't really matter. The statement:

For all [itex]\varepsilon>0[/itex], there exists an N such that for all [itex]n\geq N[/itex] holds that [itex]|a_n-a|<\varepsilon[/itex].

is actually equivalent with

For all [itex]\varepsilon>0[/itex], there exists an N such that for all [itex]n> N[/itex] holds that [itex]|a_n-a|<\varepsilon[/itex].

So you can use both statements to define limit of a sequence. Of course, once you decided on which of both versions to use, you have to be consistent.