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Series again

  • Thread starter broegger
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  • #1
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Okay, I have this function defined as an infinite series:

[tex]f(x) = \sum_{n=1}^{\infty}\frac{\sin(nx)}{x+n^4}[/tex]​

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 [tex]f(x) \rightarrow 0[/tex] as [tex]x \rightarrow \infty.[/tex] 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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Answers and Replies

  • #2
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set [tex]f_n(x) = \frac{sin(nx)}{x+n^4}[/tex], then [tex]|f_n(x)| \leq \frac{1}{n^4}[/tex] for all x>0
 
  • #3
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Hmm.. How does that show that f(x) -> 0 as x -> infinity?
 
  • #4
lurflurf
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broegger said:
Hmm.. How does that show that f(x) -> 0 as x -> infinity?
by the comparison test
 
  • #5
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lurflurf said:
by the comparison test
Maybe I'm a little slow here, but [tex]\sum_{n=1}^{\infty}\frac1{n^4}[/tex] does not equal 0, so how come?
 
  • #6
lurflurf
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broegger said:
Maybe I'm a little slow here, but [tex]\sum_{n=1}^{\infty}\frac1{n^4}[/tex] does not equal 0, so how come?
My mistake. Have you any theorems concerning the interchange of limits for uniformly convergent series?
 
  • #7
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No, I can't find any relevant theorems to apply. I'm lost at sea.
 
  • #8
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fourier jr said:
set [tex]f_n(x) = \frac{sin(nx)}{x+n^4}[/tex], then [tex]|f_n(x)| \leq \frac{1}{n^4}[/tex] 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,

Nenad
 
  • #9
CarlB
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[tex]f(x) = \frac{sin(x)}{x+1} + \sum_{n=2}^{\infty}\frac{\sin(nx)}{x+n^4}[/tex]

This is clearly bounded by:

[tex]\frac{1}{x+1} + \int_{n=1}^\infty \frac{1}{x+n^4}\,\,dn[/tex]

which can be shown to go to zero.

Carl
 
  • #10
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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
CarlB
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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
 

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