Does ln(n)sin(n) converge or diverge?

  • Thread starter Thread starter nickclarson
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around the convergence or divergence of the series \(\sum_{n=1}^{\infty} \ln(n) \sin\left(\frac{1}{n}\right)\). Participants are exploring the behavior of the terms as \(n\) approaches infinity and considering various comparison tests.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are considering the comparison test and the integral test, questioning the behavior of \(\sin\left(\frac{1}{n}\right)\) as \(n\) becomes large. There is discussion about approximating \(\sin\left(\frac{1}{n}\right)\) and comparing it to \(\frac{1}{n}\) and \(\frac{\ln(n)}{n}\).

Discussion Status

The conversation is ongoing, with participants sharing insights about the relationships between the functions involved and exploring the implications of their comparisons. Some guidance has been provided regarding the use of series expansions and the behavior of functions near zero.

Contextual Notes

Participants are grappling with the limits and behaviors of logarithmic and trigonometric functions as \(n\) approaches infinity, and there is an acknowledgment of the challenges in applying the comparison test effectively.

nickclarson
Messages
31
Reaction score
0
Does ln(n)sin(1/n) converge or diverge?

Homework Statement



[tex]\sum_{n=1}^{\infty}ln(n)sin\frac{1}{n}[/tex]

Homework Equations



-

The Attempt at a Solution



Not even sure where to start... was thinking comparison test, but if you choose [tex]b_{n} = ln(n)[/tex] you end up with [tex]\lim_{x\rightarrow\infty} sin\frac{1}{n} = 0[/tex]

which doesn't work because [tex]b_{n}[/tex] diverges... hmm.
 
Last edited:
Physics news on Phys.org
That looks like in intermediate form (inf*0).
I would suggest about finding the limit of whole thing as n -->inf

I saw this in one of my calc old exams; they used m=1/n and then did something
 
I would be thinking about a comparison to another series that I could apply the integral test to. For a small value of x, sin(x) is almost x. Does that suggest anything?
 
then I would say something like:

[tex]n\rightarrow\infty \; sin\frac{1}{n} = \frac{1}{n}[/tex]

then I can do the integral test of:

[tex]b_{n} = \frac{ln(n)}{n}[/tex]

I could be way off though, but is that what you were getting at?
 
No, that's it. That's ALMOST what you want to compare to. Now does it diverge or converge? Which is larger ln(n)*sin(1/n) or ln(n)/n?
 
[tex]\frac{lnn}{n}[/tex]

is smaller, so if it converges so does

[tex]ln(n)sin\frac{1}{n}[/tex]

that correct? Otherwise it's the other way around. What's the best method for deciding which is smaller?
 
ln n/n > 1/n
oops..nvm

but 1/n > sin (1/n)
 
It's really the same question as which is larger sin(x) or x for x near 0. Look at the graphs, punch some numbers into convince yourself. The easiest way to really answer it is to look at the series expansion of sin(x) near 0.
 
x is above sin as it approaches 0 from the positive side. However when I plug sin(1/x)ln(x) and ln(x)/x into my calculator ln(x)/x is always below... hmm
 
  • #10
If n is large, 1/n is small. That's why I suggested looking at x and sin(x) for x SMALL. If you want to directly compare sin(1/n)ln(n) and ln(n)/n, you should put n LARGE.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 34 ·
2
Replies
34
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K