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

[nickclarson]

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

Homework Statement

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

Homework Equations

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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.

 

[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

 

[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?

 

[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?

 

[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?

 

[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?

 

[rootX]

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

but 1/n > sin (1/n)

 

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

 

[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

 

[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.