Limit comparison test is applicable when `suma_n` and `sumb_n` are series with positive terms. If `lim_(n->oo)a_n/b_n=L` where L is a finite number and `L>0` , then either both series converge or both diverge.

Given series is `sum_(n=1)^oon^(k-1)/(n^k+1),k>=2`

Let the comparison series be `sum_(n=1)^oo1/n`

The comparison series `sum_(n=1)^oo1/n` is a p-series with p=1.

As per the p-series test,`sum_(n=1)^oo1/n^p` is convergent if `p>1` and divergent if `0<p<=1`

So, the comparison series is a divergent series.

Now let's use the limit comparison test with: `a_n=n^(k-1)/(n^k+1)`

`b_n=1/n`

`a_n/b_n=(n^(k-1)/(n^k+1))/(1/n)`

`a_n/b_n=(n(n^(k-1)))/(n^k+1)`

`a_n/b_n=n^k/(n^k+1)`

`a_n/b_n=n^k/(n^k(1+1/n^k))`

`a_n/b_n=1/(1+1/n^k)`

`lim_(n->oo)a_n/b_n=lim_(n->oo)1/(1+1/n^k)`

`=1>0`

Since the comparison series `sum_(n=1)^oo1/n` diverges, the series `sum_(n=1)^oon^(k-1)/(n^k+1)` as well, diverges as per the limit comparison test.