Convergence of Natural Log function with the limit comparison test

[fakecop]

Homework Statement

Determine whether Ʃ(n from 1 to infinity) ln(n)/n^3 converges or diverges using the limit comparison test.

Homework Equations

I must use the limit comparison test to solve this problem-not allowed to use other tests.

The Attempt at a Solution

I know that the series converges, because the integral test shows that the the integral does converge to a specific value. This proves that Ʃ(n from 1 to infinity) ln(n)/n^3 converges. However, I have to solve this problem with the limit comparison test.

Now my best results so far is that the rate of increase of the natural log function is slower than the rate of increase of any power function with the exponent greater than 0, so I believe that the series ln(n)/n^3 mimics the series 1/n^3. This series, 1/n^3, is the p-series whose convergence can easily be determined by looking at the exponent. If n>1, the series converges. If n≤1, the series diverges.

The same principle applies to the series ln(n)/n^k: It appears that the series converges whenever k>1 and diverges when k≤1. But how can the limit comparison test be employed? There just seems to be no suitable function to compare ln(n)/n^3 with. Please help?

 

[mfb]

There just seems to be no suitable function to compare ln(n)/n^3 with.

There is, and you posted it already:

Now my best results so far is that the rate of increase of the natural log function is slower than the rate of increase of any power function with the exponent greater than 0

Just use a power function with a (small) exponent greater than 0.

 

[STEMucator]

Homework Statement

Determine whether Ʃ(n from 1 to infinity) ln(n)/n^3 converges or diverges using the limit comparison test.

Homework Equations

I must use the limit comparison test to solve this problem-not allowed to use other tests.

The Attempt at a Solution

I know that the series converges, because the integral test shows that the the integral does converge to a specific value. This proves that Ʃ(n from 1 to infinity) ln(n)/n^3 converges. However, I have to solve this problem with the limit comparison test.

Now my best results so far is that the rate of increase of the natural log function is slower than the rate of increase of any power function with the exponent greater than 0, so I believe that the series ln(n)/n^3 mimics the series 1/n^3. This series, 1/n^3, is the p-series whose convergence can easily be determined by looking at the exponent. If n>1, the series converges. If n≤1, the series diverges.

The same principle applies to the series ln(n)/n^k: It appears that the series converges whenever k>1 and diverges when k≤1. But how can the limit comparison test be employed? There just seems to be no suitable function to compare ln(n)/n^3 with. Please help?

You have kind of the right idea so far. I don't agree though that $\frac{1}{n^3}$ behaves like $a_n = \frac{ln(n)}{n^3}$ accurately enough to determine convergence.

Notice that $ln(n) < n$ for all $n>0$. So that $\frac{ln(n)}{n^3} < \frac{n}{n^3} = \frac{1}{n^2} = b_n$.

$b _n$ clearly converges by p-series and it behaves enough like $a_n$ that you can use the limit comparison test.

What is the result of $lim_{n→∞} \frac{a_n}{b_n}$?

 

[fakecop]

There is, and you posted it already:

Just use a power function with a (small) exponent greater than 0.

Power functions will not work with the natural log function in this case (with the limit comparison test) because the limit will either be 0 or infinity, making the test inconclusive.

suppose I compare ln(n)/n^3 with 1/n^2.99. Then, I will have to evaluate the limit of ln(n)/n^0.01 as n approaches infinity, which approaches 0.

What happens when we change that 2.99 to 3? well, then I'd have to evaluate the limit of ln(n) as n approaches infinity, which approaches infinity.

The Natural log function seems to be beyond the "power" of power functions.

You have kind of the right idea so far. I don't agree though that $\frac{1}{n^3}$ behaves like $a_n = \frac{ln(n)}{n^3}$ accurately enough to determine convergence.

Notice that $ln(n) < n$ for all $n>0$. So that $\frac{ln(n)}{n^3} < \frac{n}{n^3} = \frac{1}{n^2} = b_n$.

$b _n$ clearly converges by p-series and it behaves enough like $a_n$ that you can use the limit comparison test.

What is the result of $lim_{n→∞} \frac{a_n}{b_n}$?

I believe the value of $lim_{n→∞} \frac{a_n}{b_n}$ is equal to $lim_{n→∞} \frac{ln(n)}{n}$, which equals 0-making the test inconclusive.

Now, I understand you may be trying to say that even though the test is inconclusive, because the limit is 0 (instead of infinity) and ${b_n}$ is a convergent series, the series ln(n)/n^3 is convergent. However, I need to do a limit comparison test where the test is actually conclusive. On a side note, however, ${b_n}$ does work beautifully to show that ln(n)/n^3 converges by the basic comparison test.

 

[mfb]

Power functions will not work with the natural log function in this case (with the limit comparison test) because the limit will either be 0 or infinity, making the test inconclusive.

This is wrong.

suppose I compare ln(n)/n^3 with 1/n^2.99. Then, I will have to evaluate the limit of ln(n)/n^0.01 as n approaches infinity, which approaches 0.

That is a good start. If 1/n^2.99 converges*, ln(n)/n^3 will converge, too (as the ratio between ln(n)/n^3 and 1/n^2.99 vanishes in the limit of n to infinity).
Now you just have to show that 1/n^2.99 converges.

What happens when we change that 2.99 to 3?

Well that is a bad idea.*edit for clarity: I always mean the corresponding series, of course

 

[STEMucator]

Power functions will not work with the natural log function in this case (with the limit comparison test) because the limit will either be 0 or infinity, making the test inconclusive.

suppose I compare ln(n)/n^3 with 1/n^2.99. Then, I will have to evaluate the limit of ln(n)/n^0.01 as n approaches infinity, which approaches 0.

What happens when we change that 2.99 to 3? well, then I'd have to evaluate the limit of ln(n) as n approaches infinity, which approaches infinity.

The Natural log function seems to be beyond the "power" of power functions.
I believe the value of $lim_{n→∞} \frac{a_n}{b_n}$ is equal to $lim_{n→∞} \frac{ln(n)}{n}$, which equals 0-making the test inconclusive.

Now, I understand you may be trying to say that even though the test is inconclusive, because the limit is 0 (instead of infinity) and ${b_n}$ is a convergent series, the series ln(n)/n^3 is convergent. However, I need to do a limit comparison test where the test is actually conclusive. On a side note, however, ${b_n}$ does work beautifully to show that ln(n)/n^3 converges by the basic comparison test.

Hmm that's a tough one if you want the limit to be defined as such : $0 < L < ∞$.

It can be forced to happen though within reason. We know $ln(n) < ln(n^2)$ so that $\frac{ln(n)}{n^3} < \frac{ln(n^2)}{n^3}$ and both are well behaved.

Testing $lim_{n→∞} \frac{a_n}{b_n}$ would yield a result of $L = 2$.

 

[fakecop]

That is a good start. If 1/n^2.99 converges*, ln(n)/n^3 will converge, too (as the ratio between ln(n)/n^3 and 1/n^2.99 vanishes in the limit of n to infinity).
Now you just have to show that 1/n^2.99 converges.

The ratio of ln(n)/n^3 and 1/n^2.99 approaches 0 as n approaches infinity. The limit comparison test states that the test is inconclusive if the limit is infinity. Again, your reasoning is valid (since 1/n^2.99 is convergent and the limit is 0, ln(n)/n^3 must diverge). However, it violates the rule for the limit comparison test. I have to use the limit comparison test where $0<L</infty$

Hmm that's a tough one if you want the limit to be defined as such : $0 < L < ∞$.

It can be forced to happen though within reason. We know $ln(n) < ln(n^2)$ so that $\frac{ln(n)}{n^3} < \frac{ln(n^2)}{n^3}$ and both are well behaved.

Testing $lim_{n→∞} \frac{a_n}{b_n}$ would yield a result of $L = 2$.

You're right, but then I'd have to prove that $\frac{ln(n^2)}{n^3}$ converges. But I guess there are other tests to accomplish this. This is a sneaky way to "use" the limit comparison test in the problem.:rofl:

 

[STEMucator]

You're right, but then I'd have to prove that $\frac{ln(n^2)}{n^3}$ converges. But I guess there are other tests to accomplish this. This is a sneaky way to "use" the limit comparison test in the problem.:rofl:

The question is asking you to find out about the convergence of the given series $\frac{ln(n)}{n^3}$ using the limit form of the comparison test.

It never said anything about the tests you use to confirm the convergence of $b_n$ I believe. So I think any test is fair game in that sense. How else are you going to prove $b_n$ converges without any other tests?

You're still using the limit comparison test in the end.

 

[mfb]

I have to use the limit comparison test where $0<L<\infty$

0<L is an absolutely pointless requirement.
Taken literally, the limit comparison test cannot show the convergence of anything on its own - you can just compare series with other series where you know the limit. In this class of series, the function you have to analze is the easierst one.
You can compare it with something else, but it does not make the proof of convergence easier.