Convergent Series: sin(kx)/ln(k)

[ForMyThunder]

Homework Statement

Is this series convergent for all real x:

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

Homework Equations

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.

 

[Hurkyl]

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

 

[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?

 

[Hurkyl]

Then there are at most 2pi/x terms

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.

 

[JG89]

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

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