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## Main Question or Discussion Point

It dosen't seem intuitive that if the ratio of the last two consecutive terms is less than 1 then it is convergent and viceversa if divergent when adding an infinite number of terms.

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It dosen't seem intuitive that if the ratio of the last two consecutive terms is less than 1 then it is convergent and viceversa if divergent when adding an infinite number of terms.

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quasar987

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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.

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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!

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mathwonk

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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.

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Gib Z

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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.

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The ratio test is more than just nth term as n gets large becoming 0. However its not just the n+1 term is smaller than the nth term either. Since 1/n certainly fit both descriptions. The unintuitive thing about the ratio test is that it demands the evaluation of a limit in that test.

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.

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That is nice. You are saying if the ratio test applies than this series is convergent. What about the other way? Can we apply the same argument in reverse so the proof you gave works both ways?

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.

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I was thinking about proving that if the series is convergent than the ration test applies, namely a<1.

Just before doing that I think that the series in post 2 is wrong and should look like this instead:

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

If a series converges than make it smaller then a finite series + a geometric series. Now apply the argument in reverse and get the ratio test with a<1. So it does work the other way when a<1.

It can be clearly seen that when a=1 than we can't use a geometric series so this method doesn't work.

Just before doing that I think that the series in post 2 is wrong and should look like this instead:

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

If a series converges than make it smaller then a finite series + a geometric series. Now apply the argument in reverse and get the ratio test with a<1. So it does work the other way when a<1.

It can be clearly seen that when a=1 than we can't use a geometric series so this method doesn't work.

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quasar987

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This is not cenessarily true. See below, it is b that goes in there instead of a.I was thinking about proving that if the series is convergent than the ration test applies, namely a<1.

Just before doing that I think that the series in post 2 is wrong and should look like this instead:

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

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.

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- #11

mathwonk

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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.

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

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quasar987

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

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indeed

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