Determine whether the series converges or diverges problem

[mattmannmf]

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 I am 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

 

[lanedance]

how about the magnitude of the nth term - does it go to zero for large n?

 

[Dick]

Can you show that if x close enough to 0 then sin(x)>x/2?

 

[Dick]

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).

 

[mattmannmf]

im not sure what you mean by that

 

[Dick]

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.

 

[mattmannmf]

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

 

[mattmannmf]

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?

 

[rs1n]

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 \sin(x) where x=\frac{3}{4n}. How does \sin(x) compare to \frac{x}{2} if |x|<1?

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

\sum_{n=1}^\infty \underbrace{\sin\left(\frac{3}{4n}\right)}_{\sin(x)}<br /> \quad \text{vs}\quad \sum_{n=1}^\infty \underbrace{\frac{1}{2}\cdot \frac{3}{4n}}_{\frac{1}{2}x}

 

[Dick]

Dick is suggesting that you try to compare your series with another series. Your given series has terms of the form \sin(x) where x=\frac{3}{4n}. How does \sin(x) compare to \frac{x}{2} if |x|&lt;1?

If you can show \sin(x) &gt; \frac{x}{2}, what does it say about

\sum_{n=1}^\infty \underbrace{\sin\left(\frac{3}{4n}\right)}_{\sin(x)}<br /> \quad \text{vs}\quad \sum_{n=1}^\infty \underbrace{\frac{1}{2}\cdot \frac{3}{4n}}_{\frac{1}{2}x}

Thank you for explaining it so well rs1n!

 

[mattmannmf]

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?

 

[rs1n]

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 \sum_{n=1}^\infty \left(\frac{1}{2}\cdot \frac{3}{4n}\right) converge or diverge? (What special series does it almost look like?)

 

[mattmannmf]

looks like the harmonic p-series 1/n which diverges.

 

[mattmannmf]

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?

 

[rs1n]

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.

 

[mattmannmf]

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