# Proof of the ratio test

1. Nov 7, 2012

### Bipolarity

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 : r<R<1, and then shows that there exists a N>0 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
$a_{N+1}<a_{n}R$ 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

2. Nov 7, 2012

### 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.

3. Nov 7, 2012

### 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

4. Nov 7, 2012

### micromass

It doesn't really matter. The statement:

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

is actually equivalent with

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

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.

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