How do you use the limit comparison test to determine if $Sigma (5n-3)/(n^2-2n+5)$ from $[1,oo)$ is convergent or divergent?

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Answer 1 · 2016-11-30T13:06:15

$sum_(n=1)^oo (5n-3)/(n^2-2n+5)$ is divergent

$(5n-3)/(n^2-2n+5+7n+5/4) < (5n-3)/(n^2-2n+5)$ but

$(5n-3)/(n^2-2n+5+7n+5/4) = (5n-3)/(n+5/2)^2=5/(n+5/2)-(31/2)/(n+5/2)^2$

We know that $sum_(n=1)^oo1/n^2$ is convergent and $1/(n+5/2)^2 < 1/n^2$ so

$sum_(n=1)^oo(31/2)/(n+5/2)^2$ is convergent. Now examining

$sum_(n=1)^oo5/(n+5/2) = sum_(n=3)^oo(5/(n+1/2)) gt 5sum_(n=4)^oo1/n$

and we know that $sum_(k=4)^oo 1/n$ is divergent.

Concluding

$sum_(n=1)^oo (5n-3)/(n^2-2n+5)$ is divergent

How do you use the limit comparison test to determine if Sigma (5n-3)/(n^2-2n+5)Sigma (5n-3)/(n^2-2n+5) from [1,oo)[1,oo) is convergent or divergent?