# Convergent Series

1. Jul 13, 2010

### ForMyThunder

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

Is this series convergent for all real x:

$$\sum$$$$^{\infty}_{k=2}$$$$\frac{sin(kx)}{ln(k)}$$

2. Relevant equations
3. The attempt at a solution

This series is less than

$$\frac{1}{ln(2)}$$$$\sum$$$$^{\infty}_{k=2}$$sin(kx)

which is less than $$\frac{\pi}{x ln2}$$. So, the series is bounded for all x. I'm thinking that the Dirichlet Test would show that this series converges.

Last edited: Jul 13, 2010
2. Jul 13, 2010

### Hurkyl

Staff Emeritus
Er, but sin(kx) doesn't converge to zero as k goes to infinity.

3. Jul 13, 2010

### ForMyThunder

But the sum of the sin(kx) would still be bounded, right?

If you mean the sum of sin(kx), I found that it was less than 2pi/x in this way:

Consider the interval [0,2pi] with x in this interval. Then there are at most 2pi/x terms, with some of them being < 0. Since sin(y) <= 1 for all y, we have that the sum of the |sin(kx)| < 2pi/x.

Since the partial sums of sin(kx) is bounded and 1/ln(k) is a nonincreasing nullsequence, would this mean that the the series converges by Dirichlet's Test?

Last edited: Jul 13, 2010
4. Jul 13, 2010

### Hurkyl

Staff Emeritus
The sum you wrote is over all k, not just those k for which $0 \leq kx \leq 2\pi$.

If x is not a rational multiple of 2pi, I'm not sure if I expect the sequence of partial sums of sin(kx) to be bounded or not. But if that sequence is bounded, then your argument is valid.

5. Jul 14, 2010

### JG89

Take a look at Example 1, under corollary 6, in this PDF: http://people.oregonstate.edu/~peterseb/mth311/docs/311abel.pdf [Broken]

And yes you're right, $$\sum sin(kx)$$ is bounded.

Last edited by a moderator: May 4, 2017