Question

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?

Answer 1

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

Explanation:

`(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