Limit comparison test intuition

[chipotleaway]

If we have two sequences and the ratio of their limit is greater than zero, why does this mean that they either both converge or diverge? I don't understand why the test works.

Also, what about lim[(1/x)/(1/x^2)] = lim x = ∞?

The series of 1/x^2 converges but series of 1/x diverges...

 

[pbuk]

The series of 1/x^2 converges but series of 1/x diverges...

No, both series converge on 0. Perhaps you are thinking of the sums of those series?

 

[Mark44]

If we have two sequences and the ratio of their limit is greater than zero, why does this mean that they either both converge or diverge?

If the limit is also less than infinity, then both series either converge or diverge.

The idea is that for the general term in each series, the ratio is a constant, so both series are comparable in their behavior.

I don't understand why the test works.

Also, what about lim[(1/x)/(1/x^2)] = lim x = ∞?

The series of 1/x^2 converges but series of 1/x diverges...

Note that the limit is NOT smaller than ∞, so this test does not apply.

 

[Mark44]

No, both series converge on 0.

Not true. Both sequences converge to zero, but the two series behave entirely differently.

Perhaps you are thinking of the sums of those series?

A series IS a sum.

 

[pbuk]

Removed cross post.

 

[pbuk]

Not true. Both sequences converge to zero, but the two series behave entirely differently.

A series IS a sum.

Oops, the OPs confusion is contagious! I'll get my coat...

 

[chipotleaway]

If the limit is also less than infinity, then both series either converge or diverge.

The idea is that for the general term in each series, the ratio is a constant, so both series are comparable in their behavior.

So for example if [itex]\lim_{x \to \infty} \frac{a_n}{b_n} = 10[/itex], then that's saying that 'at infinity', the terms of [itex]a_n[/itex] are 'roughly' 10 times larger than the terms of [itex]b_n[/itex] (getting closer to exactly 10 as n→∞)?

And then if [itex]\sum a_n[/itex] converges, then the [itex]\sum b_n[/itex] must converge because the terms of [itex]a_n[/itex] are larger at the 'tail end' where it matters.

Is that what you mean?

Note that the limit is NOT smaller than ∞, so this test does not apply.

Ah, so you can only use this test if the limit of the ratio of the two sequences is finite? Or does that rule only apply when one sequence is known to be convergent?

e.g. pretend we're unsure if [itex]\frac{1}{x^3}[/itex] converges or diverges but we do know that [itex]\frac{1}{x^2}[/itex]. Then we can't use the ratio test on the two because one is known to be convergent - right?

 

[Mark44]

If the limit is also less than infinity, then both series either converge or diverge.

The idea is that for the general term in each series, the ratio is a constant, so both series are comparable in their behavior.

So for example if [itex]\lim_{x \to \infty} \frac{a_n}{b_n} = 10[/itex], then that's saying that 'at infinity', the terms of [itex]a_n[/itex] are 'roughly' 10 times larger than the terms of [itex]b_n[/itex] (getting closer to exactly 10 as n→∞)?

Yes.

And then if [itex]\sum a_n[/itex] converges, then the [itex]\sum b_n[/itex] must converge because the terms of [itex]a_n[/itex] are larger at the 'tail end' where it matters.

Is that what you mean?

Not really. What you're describing sounds more like the direct comparison test than the limit comparison test. If you have a series whose terms are larger (in the tail) than those of a series that is known to diverge, then the series you're investigating also diverges. OTOH, if you have a series whose terms are smaller than those of a series that is known to converge, then your series also converges.

There are two cases where direct comparison doesn't tell you anything:
1) When the terms in the series you're testing are larger than those of a series that is known to converge.
2) When the terms in the series you're testing are smaller than those of a series that is known to diverge.

Ah, so you can only use this test if the limit of the ratio of the two sequences is finite?

No.

Or does that rule only apply when one sequence is known to be convergent?

No, the series involved can be convergent or divergent. If the limit of the ratio of the general terms is a constant, then both series have the same behavior. I.e., either both converge or both diverge.

e.g. pretend we're unsure if [itex]\frac{1}{x^3}[/itex] converges or diverges but we do know that [itex]\frac{1}{x^2}[/itex]. Then we can't use the ratio test on the two because one is known to be convergent - right?

This thread isn't about the ratio test; it's about the limit comparison test. The ratio test involves the ratio of two successive terms of a given series. The limit comparison test involves terms from two different series.

In any case, the limit comparison test doesn't apply here, not because one series is known to be convergent, but because the limit of the ratio is either 0 or ∞, depending respectively on whether your limit is (1/x³) / (1/x²) or (1/x²) / (1/x³).

 

[johnqwertyful]

The idea is that

lim an/bn=c
Then " an=c*bn " for large n so they both have the same behavior