# Determine whether the series converges or diverges problem

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

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

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

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).

im not sure what you mean by that

Dick
Homework Helper

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.

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

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?

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

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

Dick
Homework Helper

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

$$\sum_{n=1}^\infty \underbrace{\sin\left(\frac{3}{4n}\right)}_{\sin(x)} \quad \text{vs}\quad \sum_{n=1}^\infty \underbrace{\frac{1}{2}\cdot \frac{3}{4n}}_{\frac{1}{2}x}$$
Thank you for explaining it so well rs1n!

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?

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?)

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

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?

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.

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