Are these 100 series enough?

[nasi112]

Stuck at number 4. I don't understand his answer. Would the answer change if we started at n = 0?

 

[blamocur]

You are more likely to get help if you provide more information instead of expecting us to watch 6 hour video.
Here what would be helpful:

• Time code (22:25), or, even better, link to the place in the video which you refer to.
• The actual formula which you strugling with: $$\sum_{n=2}^\infty\frac{1}{(\ln n)^{\ln n}}$$

 

[blamocur]

Would the answer change if we started at n = 0?

What would be the values of $\frac{1}{(\ln n)^{\ln n}}$ for $n=0,1$ ?

 

[blamocur]

Stuck at number 4. I don't understand his answer.

I found his answer pretty clear after the first 5 minutes: which part of it are you having difficulty with?

 

[nasi112]

You are more likely to get help if you provide more information instead of expecting us to watch 6 hour video.
Here what would be helpful:

• Time code (22:25), or, even better, link to the place in the video which you refer to.
• The actual formula which you strugling with: $$\sum_{n=2}^\infty\frac{1}{(\ln n)^{\ln n}}$$

Apologies.

 

[nasi112]

What would be the values of $\frac{1}{(\ln n)^{\ln n}}$ for $n=0,1$ ?

It was a typo. I meant at n = 2. The guy in the video changed n = 1619.

$$\sum_{n=2}^{\infty}\frac{1}{(\ln n)^{\ln n}}$$ $$\sum_{n=1619}^{\infty}\frac{1}{(\ln n)^{\ln n}}$$
Do these two series have the same answer?

 

[blamocur]

Do these two series have the same answer?

They do have the same boolean answer, i.e., they both either converge or both diverge:
$$\sum_{n=2}^\infty x_n = \sum_{n=2}^{1618} x_n + \sum_{n=1619}^\infty x_n$$

 

[Dr.Peterson]

Stuck at number 4. I don't understand his answer.

Do these two series have the same answer?

I haven't bothered to watch the video. Could you at least tell us what the question is?

In particular, is it about values of sums, or about convergence of series?

Assuming it's the latter, do you see why the initial value would not matter?

 

[blamocur]

I haven't bothered to watch the video. Could you at least tell us what the question is?

In particular, is it about values of sums, or about convergence of series?

Assuming it's the latter, do you see why the initial value would not matter?

I watched some of it, and I believe the question is about convergence, not about the value of the sum (yes, the sum does converge). But I still don't understand what kind of help the OP needs.

 

[nasi112]

I found his answer pretty clear after the first 5 minutes: which part of it are you having difficulty with?

The discovery of $\frac{1}{n^2}$ to use the comparaison test is hard to find. An average student would never find that. My initial thought was to use $\frac{1}{\ln n}$. Apparently it didn't work. After watching the video multiple times, I understood what the guy did. For an average student, is there another way to solve the problem?

 

[blamocur]

The discovery of $\frac{1}{n^2}$ to use the comparaison test is hard to find. An average student would never find that. My initial thought was to use $\frac{1}{\ln n}$. Apparently it didn't work. After watching the video multiple times, I understood what the guy did. For an average student, is there another way to solve the problem?

Not all problems are made to be easy for an average student. I agree that this particular one isn't trivial, and, to be honest, I had no clue how to solve it until I watched the video.

 

[Dr.Peterson]

I still haven't looked, now because the goal is to see if a student who hasn't see the video could work it out.

I first thought of the root test, and if my scratchings are correct, then that seems to work (with a little L'Hopital thrown in).

But I agree that I wouldn't expect every such problem to be relatively easy, or for it to be at all obvious what method to try. You just try a few things and see what helps.

 

[nasi112]

Are you suggesting $\displaystyle \lim_{n \to \infty}\sqrt[\ln n]{\frac{1}{(\ln n)^{\ln n}}}$?

 

[Dr.Peterson]

No. What does the root test say?

Calculus II - Root Test

In this section we will discuss using the Root Test to determine if an infinite series converges absolutely or diverges. The Root Test can be used on any series, but unfortunately will not always yield a conclusive answer as to whether a series will converge absolutely or diverge. A proof of...

tutorial.math.lamar.edu

 

[nasi112]

No. What does the root test say?

Calculus II - Root Test

In this section we will discuss using the Root Test to determine if an infinite series converges absolutely or diverges. The Root Test can be used on any series, but unfortunately will not always yield a conclusive answer as to whether a series will converge absolutely or diverge. A proof of...

tutorial.math.lamar.edu

That what I did. I can't tell the difference.

 

[Dr.Peterson]

How is what you wrote an nth root?

 

[blamocur]

Here is a (hopefully) self-contained summary of the video segment:
Question: does this infinite series converge?:
$$\sum_{n=2}^\infty \frac{1}{(\ln n)^{\ln n}}$$Answer: yes, the series converges.
Proof: for large enough $n$ we have $a_n={(\ln n)^{\ln n}} > n^2$:
$$a_n = e^{\ln a_n} = e^{\ln n \ln\ln n} = \left(e^{\ln n} \right)^{\ln\ln n} = n^{\ln\ln n}$$so all we need is to have is: $$\ln\ln n > 2 \equiv \ln n > e^2 \equiv n > e^{e^2} \approx 1618,2$$

 

[nasi112]

How is what you wrote an nth root?

If I have $\displaystyle \frac{1}{(\ln n)^n}$, I will use the nth root. Isn't the idea of the root test to get rid of the power?

 

[Dr.Peterson]

If I have $\displaystyle \frac{1}{(\ln n)^n}$, I will use the nth root. Isn't the idea of the root test to get rid of the power?

Usually. But sometimes, when you don't see that any of the familiar methods exactly fits, you can try using methods in slightly different ways. And they may work.

I understood what the guy did. For an average student, is there another way to solve the problem?

This is what you were asking for: an alternate method. What he does in the video is to stretch the comparison test a bit, by trying something that is not obvious, and eventually making it work; my claim is that you can (again, if I did it right) stretch the root test a bit an make it work here, even though it won't be as direct as in the sort of problem you mention. In both cases, the theorem is still true, even though it takes extra work to apply it.

I wouldn't call either approach easy; but sometimes life works that way.

 

[nasi112]

I asked to solve the problem with another method that suits an average student. Even if your approach doesn't fit for me you can write it. I may understand it because I know the root test and I know the L'hopital. I'm still confused of how you will do it. Can you please start your approach to see what I didn't see?