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

  • Thread starter nickclarson
  • Start date
In summary, the conversation discusses whether the series ln(n)sin(1/n) converges or diverges. The individuals suggest using the comparison test and the integral test to determine if the series converges or diverges. They also discuss comparing ln(n)*sin(1/n) to ln(n)/n and suggest looking at the graphs of x and sin(x) for small values to determine which is larger. Ultimately, they conclude that ln(n)/n is larger and therefore, if it converges, so does ln(n)sin(1/n).
  • #1
nickclarson
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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.
 
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  • #2
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
 
  • #3
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?
 
  • #4
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?
 
  • #5
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?
 
  • #6
[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?
 
  • #7
ln n/n > 1/n
oops..nvm

but 1/n > sin (1/n)
 
  • #8
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.
 
  • #9
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.
 

1. What is the definition of convergence and divergence in mathematics?

In mathematics, convergence and divergence refer to the behavior of a sequence or series as the terms approach infinity. A sequence or series is said to converge if its terms approach a specific value or limit as the number of terms increases. On the other hand, a sequence or series is said to diverge if its terms do not approach a specific value or limit.

2. How do I determine if a given sequence or series converges or diverges?

To determine the convergence or divergence of a sequence or series, several tests can be used, such as the ratio test, the root test, or the comparison test. These tests compare the given sequence or series to a known sequence or series with a known convergence or divergence behavior.

3. What is the relationship between ln(n) and sin(n)?

There is no direct relationship between ln(n) and sin(n). However, both functions have different behaviors as n approaches infinity. The function ln(n) grows much slower than sin(n) as n increases, which can affect the convergence or divergence of the given expression.

4. Does the expression ln(n)sin(n) converge or diverge?

The convergence or divergence of the expression ln(n)sin(n) depends on the behavior of the terms ln(n) and sin(n) as n approaches infinity. If both terms approach zero as n increases, then the expression will converge. However, if either term approaches a non-zero value or oscillates as n increases, then the expression will diverge.

5. Are there any special cases where the expression ln(n)sin(n) converges or diverges?

Yes, there are special cases where the expression ln(n)sin(n) can converge or diverge. For example, if n=1, then the expression becomes ln(1)sin(1) which is equal to 0. In this case, the expression will converge. However, if n=π/2, then the expression becomes ln(π/2)sin(π/2) which is undefined. In this case, the expression will diverge.

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