# Limit of sin(2x) and ln(x)

1. May 28, 2009

### Mikabird

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

$$\Sigma$$ sin(2n)/(n*ln(n)^2) from n=2 to $$\infty$$ Determine if the summation converges or diverges.

2. Relevant equations

I cannot even figure out where to start. Should I use the Comparison Test?

3. The attempt at a solution

I thought that I could compare it to sin(2x)/2x but when n=2, sin(2x)/2x is too small.
Where am I messing up?

Last edited: May 28, 2009
2. May 28, 2009

### Bohrok

The comparison test among others only tests whether the sum converges. If you have to find the limit of the sequence, you need to try something else, probably with L'hopital's rule.

3. May 28, 2009

### Mikabird

oops, I meant find whether the summation converges or diverges. Thanks.

4. May 28, 2009

### Bohrok

Then I think the easiest would be finding another series with which to compare the original one and then use the integral test.

5. May 28, 2009

### Staff: Mentor

This is not true. The comparison test can be used also to test a series that diverges.

The key to using the comparison test is to have a good idea first whether a series converges or diverges.

In the following, I'm assuming that you're dealing with a series of nonnegative terms. If you are reasonably sure that $\sum a_n$ converges, and cn is the general term in a convergent series, and an <= cn, then your series converges.

On the other hand, if you believe that $\sum a_n$ diverges, and dn is the general term in a divergent series, and an >= dn, then your series diverges.

6. May 28, 2009

### Bohrok

You're right, and I've no excuse for not remembering to include that they test for divergence.

7. May 28, 2009

### Count Iblis

This is easy, because the series is absolutely convergent. By the triangle inequality, the absolute value of a summation is equal to or less than the summation of the absolute values. Then you can use that |sin(2n)| < 1. Since the summation of 1/[n ln^2(n)] converges (e.g. use the integral test), you find that the summation converges.