# Series again

1. Aug 22, 2005

### broegger

Okay, I have this function defined as an infinite series:

$$f(x) = \sum_{n=1}^{\infty}\frac{\sin(nx)}{x+n^4}$$​

which is converges uniformly and absolutely for x > 0. I have shown that f is continous and has a derivative for x > 0. Now I have to show that $$f(x) \rightarrow 0$$ as $$x \rightarrow \infty.$$ It's obvious that it is the case, but how do I prove it. I've tried putting 1/x outside the sum, but then I don't know about the remaining part. Any ideas?

Last edited: Aug 22, 2005
2. Aug 22, 2005

### fourier jr

set $$f_n(x) = \frac{sin(nx)}{x+n^4}$$, then $$|f_n(x)| \leq \frac{1}{n^4}$$ for all x>0

3. Aug 23, 2005

### broegger

Hmm.. How does that show that f(x) -> 0 as x -> infinity?

4. Aug 23, 2005

### lurflurf

by the comparison test

5. Aug 23, 2005

### broegger

Maybe I'm a little slow here, but $$\sum_{n=1}^{\infty}\frac1{n^4}$$ does not equal 0, so how come?

6. Aug 23, 2005

### lurflurf

My mistake. Have you any theorems concerning the interchange of limits for uniformly convergent series?

7. Aug 23, 2005

### broegger

No, I can't find any relevant theorems to apply. I'm lost at sea.

8. Aug 23, 2005

This series converges absolutely by the p-series test. I dont see why you have to look for any limits or to check it is decreases. If it comverges absolutely, then it converges.

Regards,

9. Aug 27, 2005

### CarlB

$$f(x) = \frac{sin(x)}{x+1} + \sum_{n=2}^{\infty}\frac{\sin(nx)}{x+n^4}$$

This is clearly bounded by:

$$\frac{1}{x+1} + \int_{n=1}^\infty \frac{1}{x+n^4}\,\,dn$$

which can be shown to go to zero.

Carl

10. Aug 27, 2005

### fourier jr

isn't that a much harder way to do it though? & judging by brogger's other posts i would say he's in the standard advanced calculus course where you'd learn the usual tests for convergence. (like comparison)

11. Aug 28, 2005

### CarlB

It's been a few decades, but I suppose that the integral test is still taught in freshman calculus. The advantage of using the integral test is that in addition to convergence, since the bounding integral goes to zero as x to infinity, you also get that the series goes to zero.

Carl