# Prove that the following series converges for any fixed value

1. Jul 5, 2009

### JG89

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

Prove that the following series converges for any fixed value of x: $$\sum_{n=2}^{\infty} \frac{ sin(nx) (-1)^n}{ln(n)}$$

2. Relevant equations

3. The attempt at a solution

I've tried using Leibniz's test and Abel's test. I had no luck. I've rewriting the summation using
"summation by parts" but I've had no luck doing that either. I haven't learned any convergence tests like the ratio test, root test, integral test, etc, so I assume I have to answer this question without knowledge of those tests.

Any suggestions?

2. Jul 5, 2009

### snipez90

Re: Jg89

Have you heard of Dirichlet's test? By the way, are you still reading through Courant?

3. Jul 5, 2009

### JG89

Re: Jg89

Never heard of Dirichlet's test. Later on I will return to studying Courant (that answers your question), so I will look it up then.

4. Jul 5, 2009

### Bohrok

Re: Jg89

What about the (direct) comparison test?

5. Jul 6, 2009

### JG89

Re: Jg89

I took a look at the Abel-Dedekind-Dirichlet Theorem.

It says that if the sum of a_n from n = 1 to infinity converges and the sequence b_n is of bounded variation, then the series of (a_n)(b_n) from n = 1 to infinity converges.

I will take the series from n = 1 to infinity of a_n = (-1)^n/ln(n), which converges by the alternating series test.

I now have to show that sin(nx) is of bounded variation. I suppose I will have to rewrite sin(nx) = sin((n-1)x + x) and then use the formula sin(a + b) = sin(a)cos(b) + cos(a)sin(b).

I will get cracking on that tomorrow.

6. Jul 6, 2009

### snipez90

Re: Jg89

You can also estimate sin(nx) using the identity sin(mx)sin(nx) = (1/2)[cos(m-n)x - cos(m+n)x] (easy proof). Also, I don't think it's too hard to prove that a function with a continuous derivative is of bounded variation.

Anyways, I don't know if this test is in Courant. I've mainly been reading the earlier stuff on convergence (i.e. first appendix), and that was why I asked about Courant.