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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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    I don't know how to proceed..2012-11-29
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    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.2012-11-29

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

since the harmonic series diverges so does our series.

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    Where did you use the limit comparison test here?2012-11-29
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    Can you explain please..2012-11-29
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    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?2012-11-29
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    I understand now thank you2012-11-29
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    Any time,@user43758....2012-11-29
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    Oh, it does diverge. In fact, the harmonic series $\,\sum\frac{1}{n}\,$ is considered by many, and I am one of those, as the "almost canonical" example of divergent series. It is very unlikely that in a basic calculus course you'd be given any other series as a first example of a divergent one.2012-11-29
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    So since it diverges then we can conclude that the serie I stated in the beginning diverges. Is that correct?2012-11-29
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    Yup, that is what that test, and I, said.2012-11-29