Limit comparison test is applicable, if `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/(n^2+1)`

We can compare the series with `sum_(n=1)^oon/n^2=sum_(n=1)^oo1/n`

The comparison series `sum_(n=1)^oo1/n` is a divergent harmonic series.

`a_n/b_n=(n/(n^2+1))/(1/n)=n^2/(n^2+1)`

`lim_(n->oo)a_n/b_n=lim_(n->oo)n^2/(n^2+1)`

`=lim_(n->oo)n^2/(n^2(1+1/n^2))`

`=lim_(n->oo)1/(1+1/n^2)`

`=1>0`

Since the comparison series `sum_(n=1)^oo1/n` diverges, so the series `sum_(n=1)^oon/(n^2+1)` diverges as well, by the limit comparison test.

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