Question

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

Answer 1

See below.

Explanation:

`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