## University Calculus: Early Transcendentals (3rd Edition)

Consider $a_n=\dfrac{1}{\sqrt n-1}$ When we increase the numerator, the value of the fraction will always increase and when we decrease the denominator, the value of the fraction will always increase. Thus, $\dfrac{1}{\sqrt n-1} \geq \dfrac{1}{\sqrt n}=\dfrac{1}{n^{1/2}}$ we get $a_n \leq \dfrac{1}{n^{1/2}}$ Here, $\Sigma_{n=1}^\infty \dfrac{1}{n^{1/2}},p=\dfrac{1}{2} \lt 1$, a divergent p-series. Thus the series diverges by the comparison test.