1. ## Check series convegence

Its me again
I have to find is this series is absolute and/or conditional convergent . I found that is conditional , but cant prove if it is absolute also.
Here is my work (sorry for bad grammar)
Code:
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I was going with the integral test here..

If i try ratio test L=lim n->inf |A(n+1)/An| i got L=lim n->inf [(6n+8)/(n+1)*ln(n)]*[(n*ln(n))/(6n+2)] where im stuck because i cant do the canceling if there is anything to cancel at all ...

I appreciate your help and patience guys .

2. ## Re: Check series convegence

Originally Posted by MirceM
Its me again
I have to find is this series is absolute and/or conditional convergent . I found that is conditional , but cant prove if it is absolute also.
Here is my work (sorry for bad grammar)
Code:
Image - TinyPic - Free Image Hosting, Photo Sharing & Video Hosting
I was going with the integral test here..

If i try ratio test L=lim n->inf |A(n+1)/An| i got L=lim n->inf [(6n+8)/(n+1)*ln(n)]*[(n*ln(n))/(6n+2)] where im stuck because i cant do the canceling if there is anything to cancel at all ...

I appreciate your help and patience guys .
You are asked to investigate the absolute convergence of the series $$\sum_{n=2}^{\infty} (-1)^{n+1} \left( 3+\frac{1}{n} \right) \frac{1}{\ln(n)}$$

This is absolutely convergent if and only if $$\sum_{n=2}^{\infty} \frac{1}{\ln(n)}$$ is absolutely convergent (explain why).

This last series can be compared to the harmonic series and doing so will allow you to conclude that it does not converge.

PS You are more likely to get help if you post your question in a form that the prospective helpers can read without using their WebFu

http://tinypic.com/view.php?pic=9h1hdy&s=8

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3. ## Re: Check series convegence

Thanks , im just not sure which terms i can cancel when sometimes i need to compare one series to another more simple.. And so i can prove that the last series doesnt converge by making the integral test ?

4. ## Re: Check series convegence

Originally Posted by MirceM
Thanks , im just not sure which terms i can cancel when sometimes i need to compare one series to another more simple.. And so i can prove that the last series doesnt converge by making the integral test ?
Use the comparison test, $\ln(x)<x$ for $x>0$, so $1/\ln(x)>1/x$ and you should know that the harmonic series $\sum_1^{\infty}1/x$ diverges. It is one of the standard series that you should know. (note that the lower limit on the summation is unimportant for the summability of a series.

(The integral test will show the divergence of $\sum_1^{\infty}1/x$, but normally you may assume you are allowed to use the knowledge of its divergence without proving it)

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