# Convergence Tests

## Homework Statement

Test for convergence

$$\sum$$ sin(1/n) / $$\sqrt{ln(n)}$$

sum from 2 to inf

## Homework Equations

Limit comparison test

if lim an / bn to inf
doesn't equal 1 you know if it converges or diverges
by limit comparison test

## The Attempt at a Solution

I've tried alot of different things for bn, I think the only way to test this series for convergence is from the limit comparison test.

I know the series converges I just can't prove it

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Dick
Homework Helper
How does sin(1/n) compare with 1/n? How about with 1/(2n)? Think integral test with comparison series like that.

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So , 1/n is always greater than sin(1/n)

If I just replace sin(1/n) with 1/n , I have a "an" that is less than the original and the new function diverges by the integral test.

So the sum is divergent by the comparison test?

Dick
Homework Helper
So , 1/n is always greater than sin(1/n)

If I just replace sin(1/n) with 1/n , I have a "an" that is less than the original and the new function diverges by the integral test.

So the sum is divergent by the comparison test?
You've got a divergent series that's greater than the original. That doesn't prove divergence. You need one that's less than the original. That's why I also suggested 1/(2n).

Dick
Homework Helper
Yes, Wolfram Alpha is quite wrong. Unless you believe 962 is the same as infinity.

Awesome :)

Thank you, I've spent way too much time thinking about that sum, lol.

Char. Limit
Gold Member
Okay, I see how the test would work replacing sin(1/n) with 1/2n making the new equation always less than the original. Then the integral test works and the sum is divergent.

Does that mean that wolfram is wrong though?

http://www.wolframalpha.com/input/?i=sum from 2 to inf (sin(1/n))/(sqrt(ln(n)))&t=macw01

When I plug the sum in it tells me it converges to 962 or so.
A good way to test if Wolfram-Alpha is lying is to click the More Digits button. If the number changes when you do that, W-A is probably lying to you.