# Limit comparison test intuition

1. Aug 27, 2013

### 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...

2. Aug 27, 2013

### MrAnchovy

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

3. Aug 27, 2013

### Staff: Mentor

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

4. Aug 27, 2013

### Staff: Mentor

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

5. Aug 27, 2013

### MrAnchovy

Removed cross post.

6. Aug 27, 2013

### MrAnchovy

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

7. Aug 28, 2013

### chipotleaway

8. Aug 28, 2013

### Staff: Mentor

Yes.
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.
No.
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.
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).

9. Aug 28, 2013

### johnqwertyful

The idea is that

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