Analyzing the Convergence of an an = ln\sqrt{n} / n Sequence

In summary, the conversation was about determining convergence or divergence of a sequence given the nth term and finding its limit if it converges. The sequence in question was an = ln\sqrt{n} / n, and the student initially thought it could be solved using L'Hopital's rule. They then realized their mistake and correctly applied the chain rule to find the derivative of the sequence. They were then asked to determine if the limit converges or diverges, and after some discussion and clarification, it was determined that the sequence converges to 0.
  • #1
Watson91
9
0

Homework Statement


I have been asked to determine convergence or divergence of a sequence given the nth term. If the sequence converges, find its limit.

an = ln[tex]\sqrt{n}[/tex] / n


Homework Equations





The Attempt at a Solution



I had thought that the sequence would follow L'Hopital's rule, but even trying to find the derivative of top has me at a loss when trying to decide if the sequence converges or diverges.
 
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  • #2
Can you show us an attempt to differentiate ln(sqrt(n))? You can either use the chain rule or it's easier if you use the properties of ln.
 
  • #3
If I'm not mistaken the derivative of the ln [tex]\sqrt{n}[/tex] = [tex]\frac{1}{2n}[/tex]

Please correct me if I'm wrong, I'm quite unsure of my work at this point to be honest.
 
  • #4
Watson91 said:
If I'm not mistaken the derivative of the ln [tex]\sqrt{n}[/tex] = [tex]\frac{1}{2n}[/tex]

Please correct me if I'm wrong, I'm quite unsure of my work at this point to be honest.

That's right. If you want to be even more sure you could show us how you derived it.
 
  • #5
Derivative = 1/n X 1/2[tex]\sqrt{n}[/tex] = [tex]\frac{1}{2n}[/tex]
 
  • #6
Watson91 said:
Derivative = 1/n X 1/2[tex]\sqrt{n}[/tex] = [tex]\frac{1}{2n}[/tex]

You didn't say how you did that or anything. Could you put yourself in my place? Suppose somebody told you "1/n X 1/2[tex]\sqrt{n}[/tex]" is equal to "[tex]\frac{1}{2n}[/tex]". Would you say that's right or would you worry about it? I know you have the right answer, but what you've provided doesn't really convince me you did it correctly.
 
  • #7
I understand your concern now. I used what I thought was the chain rule by taking the derivative of the outside ( 1/n ) and then multiplied by the derivative of in the inside ( 1/2( sqrt(n) ).

Really I did the problem incorrectly I can see because what I tried to say was n X sqrt(n) = n which isn't true. I stumbled across the right answer on accident I guess.

I think in actuality it should be derivative of outside ( 1/sqrt(n) ) x derivative of inside ( 1/2(sqrt(n)) ) which would now give me the answer of 1/2n )

Hopefully this time I'm correct and described it correctly.
 
  • #8
Watson91 said:
I understand your concern now. I used what I thought was the chain rule by taking the derivative of the outside ( 1/n ) and then multiplied by the derivative of in the inside ( 1/2( sqrt(n) ).

Really I did the problem incorrectly I can see because what I tried to say was n X sqrt(n) = n which isn't true. I stumbled across the right answer on accident I guess.

I think in actuality it should be derivative of outside ( 1/sqrt(n) ) x derivative of inside ( 1/2(sqrt(n)) ) which would now give me the answer of 1/2n )

Hopefully this time I'm correct and described it correctly.

Nice. Now it's perfect. Thanks. You could also have done it by using the rule of logs that ln(sqrt(n))=ln(n^(1/2))=(1/2)ln(n). Then take the derivative. You get the same thing, of course.
 
  • #9
Great, now back to my original question. How do I decide if the limit converges or diverges from here? This is the part I'm really stuck/confused about.
 
  • #10
Watson91 said:
Great, now back to my original question. How do I decide if the limit converges or diverges from here? This is the part I'm really stuck/confused about.

Sure you don't want to answer it yourself? The derivative of the numerator is 1/(2n). What's the limit of that as n goes to infinity? The derivative of the denominator is 1. It's l'Hopital, right?
 
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  • #11
Well I know that the lim of 1/n = 0 and diverges. By comparison I would say that 1/2n > 1/n which would make me believe it diverges. Would this be correct?
 
  • #12
Watson91 said:
Well I know that the lim of 1/n = 0 and diverges. By comparison I would say that 1/2n > 1/n which would make me believe it diverges. Would this be correct?

I think you are confusing convergence of a sequence with convergence of a series. I thought you were talking about the former. That's where you use l'Hopital. Are you summing the terms or is it just the limit of the terms?
 
  • #13
What the question asks is to determine convergence or divergence. Maybe I am confusing the two.

I think what would answer my question is; if you are able to find the limit then your sequence converges, I think this is correct for a sequence. To my knowledge the limit of 1/2n = 0, so if I'm correct there the sequence an converges then.
 
  • #14
Watson91 said:
What the question asks is to determine convergence or divergence. Maybe I am confusing the two.

I think what would answer my question is; if you are able to find the limit then your sequence converges, I think this is correct for a sequence. To my knowledge the limit of 1/2n = 0, so if I'm correct there the sequence an converges then.

Yes, the SEQUENCE a_n converges to zero. The SERIES a_n diverges. I think they are asking about the sequence, since that's what you said in the original post.
 
  • #15
Yep the question is of the sequence. Thanks for helping me out on that problem, even though I drug it on. It really wasn't that difficult after I look back over it, but it always seems that way for me. I get a mental block and once I get past it I question why I couldn't solve it originally.

Again, thanks for sticking with me and stepping me through it.
 
  • #16
You're welcome. You'll do fine once you get over 'blocking'. You are good at understanding where you went wrong. This went a lot better than a many 'drug on' exchanges I've been involved with.
 
  • #17
Yeah blocking is a problem for me when it comes to calculus. I seem to get hung up on things that shouldn't be a problem for myself. I'm not sure how to overcome the issue though other than just asking for help.
 

1. What is the formula for the given sequence?

The formula for the given sequence is an = ln√n / n.

2. How do you determine if the given sequence converges or diverges?

To determine if the sequence converges or diverges, we use the limit comparison test or the ratio test.

3. What is the limit of the given sequence?

The limit of the sequence is 0, as n approaches infinity.

4. Can you provide an example of the convergence of the given sequence?

An example of the convergence of the given sequence is when n = 10, an = ln√10 / 10 ≈ 0.1155.

5. Is the convergence of this sequence affected by the initial value of n?

No, the convergence of the sequence is not affected by the initial value of n. It is determined by the limit as n approaches infinity.

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