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

1. Apr 17, 2008

### nickclarson

Does ln(n)sin(1/n) converge or diverge?

1. The problem statement, all variables and given/known data

$$\sum_{n=1}^{\infty}ln(n)sin\frac{1}{n}$$

2. Relevant equations

-

3. The attempt at a solution

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

which doesn't work because $$b_{n}$$ diverges... hmm.

Last edited: Apr 17, 2008
2. Apr 17, 2008

### rootX

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. Apr 17, 2008

### Dick

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. Apr 17, 2008

### nickclarson

then I would say something like:

$$n\rightarrow\infty \; sin\frac{1}{n} = \frac{1}{n}$$

then I can do the integral test of:

$$b_{n} = \frac{ln(n)}{n}$$

I could be way off though, but is that what you were getting at?

5. Apr 17, 2008

### Dick

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. Apr 17, 2008

### nickclarson

$$\frac{lnn}{n}$$

is smaller, so if it converges so does

$$ln(n)sin\frac{1}{n}$$

that correct? Otherwise it's the other way around. What's the best method for deciding which is smaller?

7. Apr 17, 2008

### rootX

ln n/n > 1/n
oops..nvm

but 1/n > sin (1/n)

8. Apr 17, 2008

### Dick

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 in to convince yourself. The easiest way to really answer it is to look at the series expansion of sin(x) near 0.

9. Apr 17, 2008

### nickclarson

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. Apr 17, 2008

### Dick

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.