Answer
Diverges
Work Step by Step
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.
