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I am having some trouble with the following exercise:

I need to determine if the following serie converges or diverges using only the limit comparison test:

$\sum_{n=1}^{\infty} \frac{n}{(4n-3)(4n-1)}$

Please help.

Thank you in advance

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what have you tried? –  user127.0.0.1 Nov 29 '12 at 12:19
    
I don't know how to proceed.. –  Carpediem Nov 29 '12 at 12:19
    
If you like an answer you could upvote it; you may want to wait a while for some possible future better answers to choose it as "the best answer", but any answer that helps you a little should be, imo, upvoted. –  DonAntonio Nov 29 '12 at 12:37
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1 Answer 1

up vote 1 down vote accepted

$$\lim_{n\to\infty}\frac{\frac{n}{(4n-3)(4n-1)}}{\frac{1}{n}}=\lim_{n\to\infty}\frac{n^2}{16n^2-16n+3}=\frac{1}{16}\Longrightarrow$$

since the harmonic series diverges so does our series.

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Where did you use the limit comparison test here? –  Carpediem Nov 29 '12 at 12:21
    
Can you explain please.. –  Carpediem Nov 29 '12 at 12:25
    
Do you know the limit comparison test for positive series? It says: let $\,\sum a_n\,\,,\,\,\sum b_n\,$ be positive series s.t. $\,\lim\frac{a_n}{b_n}\,$ exists finitely. Then $\,\sum a_n\,$ converges iff $\,\sum b_n\,$ converges. What's not clear here? –  DonAntonio Nov 29 '12 at 12:28
    
I understand now thank you –  Carpediem Nov 29 '12 at 12:30
    
Any time,@user43758.... –  DonAntonio Nov 29 '12 at 12:32
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