# Math Help - is the series Sigma(1 to oo) sin/n convergent or divergent?

1. ## is the series Sigma(1 to oo) sin/n convergent or divergent?

omg
Really OMG !!
It seems easy but really i cant test it !!

Is this one really hard or am forgetting my informations about testing the series for convergence ??!

2. Originally Posted by TWiX
omg
Really OMG !!
It seems easy but really i cant test it !!

Is this one really hard or am forgetting my informations about testing the series for convergence ??!

See here: http://www.mathhelpforum.com/math-he...-n-series.html

3. Thank you.
Thanks for my god .. Its hard, I thought am loosing the informations

4. As an additional info: we know that the partial sum for $\sin n$ is bounded so $\sum a_n\sin n$ converges by Dirichlet test as long as $a_n$ is a decreasing sequence which tends to zero.

5. Actually, I dont know this test, but i think its the test with 2^n or something like that, Right?

6. No, that's related with the Condensation test.

Look at wikipedia, and seek for Dirichlet test and Cauchy's condensation test.

7. Isn't it the case that $-1\leq\sin(n)\leq1$ ?

And doesn't this imply that

$\sum_{n=1}^{\infty}\frac{\sin{n}}{n}$ can be directly compared to $\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$ which converges by the alternating series test?

8. Originally Posted by VonNemo19
Isn't it the case that $-1\leq\sin(n)\leq1$ ?

And doesn't this imply that

$\sum_{n=1}^{\infty}\frac{\sin{n}}{n}$ can be directly compared to $\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$ which converges by the alternating series test?
OK $\frac{sin(n)}{n}\geq\frac{-1}{n}$
But what about $\frac{(-1)^n}{n}$ ?
ohhh since $(-1)^n$ = $\pm 1$
Am not sure about inquality, sometimes this bigger and sometimes its smaller.
You are telling me $\frac{sin(n)}{n}\geq\frac{(-1)^n}{n}$ ?
If the smaller converges, You cant decide that convergence of the bigger.
This is what I know.
Or you are telling me $\frac{sin(n)}{n}\leq\frac{(-1)^n}{n}$ ?

9. No, that's wrong.

In order to be the Comparison test applied, the general term must be nonnegative.

10. Originally Posted by Krizalid
No, that's wrong.

In order to be the Comparison test applied, the general term must be nonnegative.
WOW.
I said that there is something wrong somewhere

But It is correct that $\frac{sin(n)}{n}\leq\frac{(-1)^n}{n}$ ??

11. Originally Posted by Krizalid
No, that's wrong.

In order to be the Comparison test applied, the general term must be nonnegative.
Alternating series...

12. No, re-read what I said.

You're using comparison test, and $a_n=\frac{\sin n}{n}$ is not always positive, so you can't use a direct comparison test.

13. Originally Posted by Krizalid

You're using comparison test, and $a_n=\frac{\sin n}{n}$ is not always positive, so you can't use a direct comparison test.
Sir I know that I cant use the comparison tests.

Is it right ?

Lol you are commented on his reply.
Sorry.
But is the inquality right?

14. No.

15. Originally Posted by TWiX
WOW.
I said that there is something wrong somewhere

But It is correct that $\frac{sin(n)}{n}\leq\frac{(-1)^n}{n}$ ??
$\frac{\sin{3}}{3} = 0.04704000$

$\frac{(-1)^3}{3} = -0.333333 < 0.0470400$

For example.

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