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

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

Homework Statement

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

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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.

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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

Dick
Homework Helper
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:

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

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?

Dick
Homework Helper
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?

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

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?

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

but 1/n > sin (1/n)

Dick
Homework Helper
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.

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

Dick