How do you prove the validity of the ratio test?

I won't prove it but I'll try to convince you that it's intuitive.

Say we have a series of general term c_n and the ratio test is conclusive. Say,

[tex]\lim_{n\rightarrow \infty} \left|\frac{c_{n+1}}{c_n}\right|=a<1[/tex]

This means that by taking n large enough, we can make |c_{n+1}| as close as we want to a*|c_n|. So, for say n>N, we can approximate |c_{N+1}| by a*|c_N|, |c_{N+2}| by a*|c_{N+1}|=a²*|c_N|, etc, such that the series looks like

[tex]\sum_{n=1}^{\infty}|c_n|=\sum_{n=1}^{N}|c_n|+|c_N|\sum_{n=1}^{\infty}a^n[/tex]

I.e. like some finite sum + a geometrical series. So the series converges absolutely.

 

quasar's post resembles how you could construct a proof.

Assume WLOG that [itex]c_n[/itex] is positive for every n. Suppose

[tex]\lim_{n \rightarrow \infty} \frac{c_{n+1}}{c_n} = a <1.[/tex]

Let [itex]\delta \in (a, 1)[/itex]. Then there's some N s.t.

[tex]|c_{n+1}/c_n - a| < \delta - a[/tex]

for all [itex]n>N[/itex]. Then for all [itex]n>N[/itex], certainly,

[tex]c_{n+1} < \delta c_n.[/tex]

See if you can finish!

 

there are only 2 convergent processes we understand easily, the geometric series, and a convergent integral.


all tests for absolute covergence consist of comparing the size of the terms to one of these.

the ratio test amounts to saying the terms are eventually bunded by those of a coinvergent geometric series.

there is no clearer explanatio of this than in courant vol 1.

 

Perhaps it seems unintuitive because You think that it compares the "last" 2 terms.

I seem to think of it as since as n gets large the terms are getting smaller, ie the ratio is less than 1, than the terms must be approaching zero. Since the terms are approaching zero, it seems intuitive that it converges.

 

The other way is quite similar. Suppose a > 1. Then you can do the same thing to bound your series from below by a geometric series with common ration a > 1, which diverges.

 

This is not cenessarily true. See below, it is b that goes in there instead of a.

In any case, post #2 was not meant to be taken too seriously. I was only trying to make you understand intuitively what it meant and what it implies for the ratio of c_{n+1} to c_n to get smaller than 1 as n gets bigger.

However, it only takes a few more technicality to make that idea rigorous. Inspired by Data's post, will try to do that now.

Ok, say that as before,

[tex]\lim_{n\rightarrow \infty} \left|\frac{c_{n+1}}{c_n}\right|=a<1[/tex]

This means that, in particular, for [itex]\epsilon \in (0,a)[/itex], there exists an N s.t. for all [itex]n\geq N[/itex],

[tex]\left|\frac{c_{n+1}}{c_n}-a\right|<\epsilon \Leftrightarrow |c_{n+1}-ac_n|<\epsilon|c_n| \Rightarrow |c_{n+1}|-a|c_n|<\epsilon |c_n| \Leftrightarrow |c_{n+1}|<(\epsilon+a)|c_n|\equiv b|c_n|[/tex]

Because I have chosen epsilon in (0,a), b is less than 1. So to recapitulate, what the above is saying is that for all [itex]n\geq N[/itex], [itex]|c_{n+1}|<b|c_n|[/itex]. In particular, [itex]|c_{N+1}|<b|c_N| \Rightarrow |c_{N+2}|<b|c_{N+1}|<b^2|c_N|\Rightarrow |c_{N+3}|<...[/itex], so we can write

[tex]\sum_{n=1}^{\infty}|c_n|=\sum_{n=1}^{N}|c_n|+\sum_{n=N+1}^{\infty}|c_n|<\sum_{n=1}^{N}|c_n|+|c_N|\sum_{n=1}^{\infty}b^n<\infty[/tex]

The second sum starts at 1 and not at N+1 because as I said in the above paragraph, |c_{N+1}| is bounded by b|c_N|, not b^{N+1}|c_N|. Anyway, I feel pretty intelligent right now. So this is what it's like to be StatusX, huh. I like it.

 

if you read my post you see there are two tests, comparing with geometric series and comparing with convergent integrals. if the ratio test always worked, there would be no need for the integral test.

a standard example is the sum of the terms 1/n^2.

here the ratio tends to 1, so the ratio test fails, but the integral of 1/x^2 converges, so the series does too.

 

quasar, I think you have a little bit of a problem, but it's an easy fix. Having [itex]\epsilon \in (0,a)[/itex] does not quite ensure that [itex]\epsilon + a < 1[/itex]. See how to fix it?

The "diverges when a>1" part is, as Moo indicated, by an almost identical argument.

 

Oops, make that [tex]\epsilon\in(0,1-a)[/tex].

 

indeed