How do you use the limit comparison test to determine if $Sigma n/(n^2+1)$ from $[1,oo)$ is convergent or divergent?

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Answer 1 · 2017-01-23T12:29:38

See below.

$n/(n^2+n) < n/(n^2+1) < n/n^2$ or

$1/(n+1) < n/(n^2+1) < 1/n$

Here $sum_(n=1)^oo1/(n+1) le sum_(n=1)^oon/(n^2+1)le sum_(n=1)^oo1/n$

but

$1+sum_(n=1)^oo1/(n+1) =sum_(n=1)^oo1/n =oo$

so $sum_(n=1)^oon/(n^2+1)$ is divergent

How do you use the limit comparison test to determine if Sigma n/(n^2+1)Sigma n/(n^2+1) from [1,oo)[1,oo) is convergent or divergent?