# Series again

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?

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

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

lurflurf
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broegger said:
Hmm.. How does that show that f(x) -> 0 as x -> infinity?
by the comparison test

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

lurflurf
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broegger said:
Maybe I'm a little slow here, but $$\sum_{n=1}^{\infty}\frac1{n^4}$$ does not equal 0, so how come?
My mistake. Have you any theorems concerning the interchange of limits for uniformly convergent series?

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

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

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

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

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)

CarlB