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Determine whether the series converges or diverges problem

  • Thread starter mattmannmf
  • Start date
  • #1
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given the series, determine whether the series converges or diverges:
(E is my sigma)

E sin(4/3n)

originally i started with the integral test and i got up to the point where
(integral) sin (4/(3x)) dx

but i could not come up with how to solve the integral yet. (maybe by parts)

I was talking to a friend and he said he solved it by direct comparison or limit comparison.
Now he said he used 1/n (harmonic p-series diverges) to relate to the problem but im stuck on how to get rid of the sin. Once i figure that out that its all easy.

Am i able to remove the sin out of the whole problem or what? not sure. any help would be great
 

Answers and Replies

  • #2
lanedance
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how about the magnitude of the nth term - does it go to zero for large n?
 
  • #3
Dick
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Can you show that if x close enough to 0 then sin(x)>x/2?
 
  • #4
Dick
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how about the magnitude of the nth term - does it go to zero for large n?
The poster means sin(4/(3n)), not sin((4/3)*n).
 
  • #5
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im not sure what you mean by that
 
  • #6
Dick
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im not sure what you mean by that
Pretty much what I said to begin with. I mean can you show that if x is positive and close to zero, that sin(x)>x/2? That's one way to 'get rid of the sin' by using a comparison test.
 
  • #7
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Can you show that if x close enough to 0 then sin(x)>x/2?
what do you mean if x get close enough to 0 than sin(x)>x/2?

are you talking about the squeeze thrm?
ill try to solve the integral using by parts
 
  • #8
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where are you getting the x?

i only got the x from doing the integral test. if not than shouldn't it be n's?
 
  • #9
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im not sure what you mean by that
Dick is suggesting that you try to compare your series with another series. Your given series has terms of the form [itex]\sin(x)[/itex] where [itex]x=\frac{3}{4n}[/itex]. How does [itex]\sin(x)[/itex] compare to [itex]\frac{x}{2}[/itex] if [itex]|x|<1[/itex]?

If you can show [itex]\sin(x) > \frac{x}{2}[/itex], what does it say about

[tex]\sum_{n=1}^\infty \underbrace{\sin\left(\frac{3}{4n}\right)}_{\sin(x)}
\quad \text{vs}\quad \sum_{n=1}^\infty \underbrace{\frac{1}{2}\cdot \frac{3}{4n}}_{\frac{1}{2}x}[/tex]
 
  • #10
Dick
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Dick is suggesting that you try to compare your series with another series. Your given series has terms of the form [itex]\sin(x)[/itex] where [itex]x=\frac{3}{4n}[/itex]. How does [itex]\sin(x)[/itex] compare to [itex]\frac{x}{2}[/itex] if [itex]|x|<1[/itex]?

If you can show [itex]\sin(x) > \frac{x}{2}[/itex], what does it say about

[tex]\sum_{n=1}^\infty \underbrace{\sin\left(\frac{3}{4n}\right)}_{\sin(x)}
\quad \text{vs}\quad \sum_{n=1}^\infty \underbrace{\frac{1}{2}\cdot \frac{3}{4n}}_{\frac{1}{2}x}[/tex]
Thank you for explaining it so well rs1n!
 
  • #11
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ok. So i understand that you substituted x= 3/4n

if i show that sin(x) > x/2 than the series converges. is that what your saying?
 
  • #12
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ok. So i understand that you substituted x= 3/4n

if i show that sin(x) > x/2 than the series converges. is that what your saying?
Look up the direct comparison test, and see if you are applying it correctly. Does the series [itex]\sum_{n=1}^\infty \left(\frac{1}{2}\cdot \frac{3}{4n}\right)[/itex] converge or diverge? (What special series does it almost look like?)
 
  • #13
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looks like the harmonic p-series 1/n which diverges.
 
  • #14
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so are you saying that by comparing 1/n p-series diverges

then (1/2* 3/4n) must also diverge

indicating that sin(3/4n) diverges also?
 
  • #15
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so are you saying that by comparing 1/n p-series diverges

then (1/2* 3/4n) must also diverge

indicating that sin(3/4n) diverges also?
That's the idea, but you actually need to show that those statements are true.
 
  • #16
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Now wouldn't it just be the same if i compared 1/n to 3/4n to diverge

Then could I just say since 3/4n diverges then sin(3/4n) must also diverge?

I wouldn't really know how to do the algebra for the direct comparison test or limit comparison test with the sin being there. But i can do the algebra for comparing 1/n with 3/4n
 

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