`sum_(n=0)^oo 1/sqrt(n^2+1) ` Use the Limit Comparison Test to determine the convergence or divergence of the series.

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=0)^oo1/sqrt(n^2+1)`

Let's compare with the series `sum_(n=1)^oo1/sqrt(n^2)=sum_(n=1)^oo1/n`

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

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

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

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

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

`=1>0`

Since the comparison series `sum_(n=1)^oo1/n` diverges, so the series `sum_(n=0)^oo1/sqrt(n^2+1)` diverges as well, by the limit comparison test.

Comments

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