# Math Help - Can't solve these 3 series questions.

1. ## Can't solve these 3 series questions.

$\Sigma_{n=2}^\infty \frac{(-1)}{\ln(n)}^2$. I have to decide if the series is abs convergent, cond conv, or div. I think it is abs convergent because if if i take the abs value i end up with a p series. Is this correct?

Also I'm not sure how to approach this problem

$\Sigma_{n=1}^{\infty} \frac{sin(\frac{1}{n})}{n}$

I know the sequence converges to 0 but not sure how to go about testing if the series converges or diverges. I thought about using the comp test but I'm not able to use it. Any suggestions would be appreciated, thanks.

2. Originally Posted by 11rdc11
$\Sigma_{n=2}^\infty \frac{(-1)}{\ln(n)}^2$. I have to decide if the series is abs convergent, cond conv, or div. I think it is abs convergent because if if i take the abs value i end up with a p series. Is this correct?

Also I'm not sure how to approach this problem

$\Sigma_{n=1}^{\infty} \frac{sin(\frac{1}{n})}{n}$

I know the sequence converges to 0 but not sure how to go about testing if the series converges or diverges. I thought about using the comp test but I'm not able to use it. Any suggestions would be appreciated, thanks.
I think the first one should diverge by the integral test

3. I pressume there's a typo in the first problem, if you meant $a_n=\frac{(-1)^n}{\ln n},$ then this doesn't converge absolutely since $\frac1{\ln n}>\frac1n,$ but by Leibniz test the series converges, so the series converges conditionally.

4. 1) Are you serious? Are you sure that exponent shouldn't be "n"?
1) Compare to 1/n.

2) Possible Hint: sin(x) < x for positive x very nearly zero.

5. Originally Posted by 11rdc11
$\Sigma_{n=2}^\infty \frac{(-1)}{\ln(n)}^2$. I have to decide if the series is abs convergent, cond conv, or div. I think it is abs convergent because if if i take the abs value i end up with a p series. Is this correct?

Also I'm not sure how to approach this problem

$\Sigma_{n=1}^{\infty} \frac{sin(\frac{1}{n})}{n}$
limit comparison w/ the known convergent series $\sum \frac{1}{n^2}$

$\lim_{n \to \infty} \frac{\frac{\sin(1/n)}{n}}{\frac{1}{n^2}} =$

$\lim_{n \to \infty} n\sin(1/n) =$

$\lim_{n \to \infty} \frac{\sin(1/n)}{\frac{1}{n}} = 1$

series converges.

6. Yes sorry the first one is $(\ln{n})^2$ and yes it I found out it diverges by the the limit comparison test by letting a_n equal 1/n

7. This really is not a useful response. Now that you have corrected the typo, what have you discovered. You have lot's of hints. What are your conclusions.

8. Originally Posted by TKHunny
This really is not a useful response. Now that you have corrected the typo, what have you discovered. You have lot's of hints. What are your conclusions.
I'm not sure what you mean by what was ntoa useful response. I found out the first series has to be cond conv because it diverges using the limit com test yet converges using the alter series test. And thanks Skeeter. I kept using the direct comparsion test and was having problems finding the right funstion to compare it too. One last question, how do I know when to use the direct comp test rather than the limit comp test. Series is the only section in Calculus that gave me problems and now it giving me problems in Diff equations.

9. All you did was reiterate information that was already known or speculated. You did not provide any information concerning any increased understanding or any willingness to pursue the solution. That is not useful.

Which test? The idea is to prove it. You use what works. Do not get lost in rules when to use what test. Some tests are more consistently helpful in some situations. No test is perfectly helpful in all situations.

10. Of course, others may also be waiting for the third one.

11. when a series converges to zero, try L'hopital rule