Why does the Limit Comparison Test fail?

[vanceEE]

Homework Statement

$$\sum\limits_{n=1}^∞ \frac{1}{n√(n)} $$

Since $$ \frac{1}{n√(n)} \equiv \frac{1}{x^{3/2}} $$ this is a convergent p-series. But, when I attempt to prove this by the limit comparison test with known convergent series such as $$\sum\limits_{n=1}^∞ \frac{1}{n^2}$$
ex. $$\lim_{n \to ∞} \frac{\frac{1}{n√(n)}}{\frac{1}{n^2}} = ∞ $$
The limit comparison test does not prove this fact. But I get a real number N when comparing the known divergent series $$\sum\limits_{n=1}^∞ \frac{1}{n}$$ with $$\sum\limits_{n=1}^∞ \frac{1}{n√(n)} $$
ex. $$\lim_{n \to ∞} \frac{\frac{1}{n√(n)}}{\frac{1}{n}} = 0 $$ Can someone please explain the reasoning behind this?

 

[Dick]

Homework Statement

$$\sum\limits_{n=1}^∞ \frac{1}{n√(n)} $$

Since $$ \frac{1}{n√(n)} \equiv \frac{1}{x^{3/2}} $$ this is a convergent p-series. But, when I attempt to prove this by the limit comparison test with known convergent series such as $$\sum\limits_{n=1}^∞ \frac{1}{n^2}$$
ex. $$\lim_{n \to ∞} \frac{\frac{1}{n√(n)}}{\frac{1}{n^2}} = ∞ $$
The limit comparison test does not prove this fact. But I get a real number N when comparing the known divergent series $$\sum\limits_{n=1}^∞ \frac{1}{n}$$ with $$\sum\limits_{n=1}^∞ \frac{1}{n√(n)} $$
ex. $$\lim_{n \to ∞} \frac{\frac{1}{n√(n)}}{\frac{1}{n}} = 0 $$ Can someone please explain the reasoning behind this?

What is the statement of the limit comparison test? Does it tell you anything useful in those two cases? Limit tests aren't guaranteed to to work. That's why there are so many of them. Sometimes the conclusion you draw is "inconclusive". Then you try another test. Like the p-series test.

 

[vanceEE]

What is the statement of the limit comparison test? Does it tell you anything useful in those two cases?

Well, from these two comparisons, I see that $$ \frac{1}{n}\leq\frac{1}{n^{3/2}}<\frac{1}{n^2} $$ (which can easily be deduced from the start), but this means that it does me something a bit useful right? It shows that the series is in fact convergent by the direct comparison test.. Thanks man!

 

[Dick]

Well, from these two comparisons, I see that $$ \frac{1}{n}\leq\frac{1}{n^{3/2}}<\frac{1}{n^2} $$ (which can easily be deduced from the start), but this means that it does me something a bit useful right? It shows that the series is in fact convergent by the direct comparison test.. Thanks man!

That's not true at all. What I believe it is telling you is the in the limit $\frac{1}{n}\gt\frac{1}{n^{3/2}}\gt\frac{1}{n^2}$. I don't think that's in any way useful.

 

[D H]

Well, from these two comparisons, I see that $$ \frac{1}{n}\leq\frac{1}{n^{3/2}}<\frac{1}{n^2} $$ (which can easily be deduced from the start), but this means that it does me something a bit useful right? It shows that the series is in fact convergent by the direct comparison test.. Thanks man!

That comparison tells you that the series might be divergent, but then again that it might be convergent.

In short, it tells you nothing useful.

 

[Dick]

That comparison tells you that the series might be divergent, but then again that it might be convergent.

In short, it tells you nothing useful.

Um, the comparison you quoted is wrong. The inequality signs are in the wrong direction.

 

[D H]

Missed that detail.

vanceEE, even after correcting your notation, (it should be $1/n > 1/n^{3/2} > 1/n^2$ for all n>1), this tells you nothing.

That $1/n>1/n^{3/2}$ tells you nothing. If it was the other way around, $1/n< a_n$, you would know you have a divergent series on hand. Similarly, that $1/n^{3/2}>1/n^2$ also tells you nothing. Once again, if it was the other way around, $1/n^2 > a_n$, you would know that the series on hand is convergent.

That the terms lie between those of a series known to be divergent and those of a series known to be convergent? There's no useful information here.

 

[Dick]

Missed that detail.

vanceEE, even after correcting your notation, (it should be $1/n > 1/n^{3/2} > 1/n^2$ for all n>1), this tells you nothing.

That $1/n>1/n^{3/2}$ tells you nothing. If it was the other way around, $1/n< a_n$, you would know you have a divergent series on hand. Similarly, that $1/n^{3/2}>1/n^2$ also tells you nothing. Once again, if it was the other way around, $1/n^2 > a_n$, you would know that the series on hand is convergent.

That the terms lie between those of a series known to be divergent and those of a series known to be convergent? There's no useful information here.

Detail? The OP had already drawn wrong conclusions based on it, and I corrected the direction. Don't you read previous posts?

 

[Mark44]

To elaborate on what D H said, suppose you are investigating the behavior of a given series. Also suppose that you have one series that is known to be convergent and another that is known to be divergent. There are four possibilities.

1. Your series is term-by-term larger than the known convergent series.
2. Your series is term-by-term smaller than the known convergent series.
3. Your series is term-by-term larger than the known divergent series.
4. Your series is term-by-term smaller than the known divergent series.

Items 1 and 4 in the list above are no help. Only items 2 and 3 offer some help. #2 guarantees that your series converges (it is smaller than a convergent series). #3 guarantees that your series diverges (it is larger than a divergent series).

 

[D H]

Detail? The OP had already drawn wrong conclusions based on it, and I corrected the direction. Don't you read previous posts?

How do you know the OP has drawn wrong conclusions from that typo? Perhaps it was just that, a typo. That's how I read it. In fact, I oftentimes don't see typos. My brain auto-corrects mathematics. I have to read character-by-character to see the math typos. I read the the OP's erroneous $1/n \le 1/n^{3/2} < 1/n^2$ as $1/n > 1/n^{3/2} > 1/n^2$. (Aside: When I write professionally I make sure that I have a good reviewer on hand to read my writing because I can't see my own typos. My brain reads what I meant to type rather than what I actually did type.)

Even after correcting for that typo, the OP was still wrong. That $1/n > 1/n^{3/2} > 1/n^2$ for all n>1 says nothing about the convergence or divergence of the series $\sum 1/n^{3/2}$.

 

[vela]

How do you know the OP has drawn wrong conclusions from that typo?

Because the OP concluded the series converged when you can't conclude anything from those comparisons.

 

[Dick]

Because the OP concluded the series converged when you can't conclude anything from those comparisons.

I can conclude that, in fact, D H doesn't read previous posts.