Answer
Diverges
Work Step by Step
Let $u_n=\dfrac{1}{\sqrt n-1}$
When the numerator increases, the value of fraction is always increases and also, when the denominator decreases, the value of fraction is always increases.
$\implies \dfrac{1}{\sqrt n-1} \geq \dfrac{1}{\sqrt n}=\dfrac{1}{n^{1/2}}$
so, $u_n \leq \dfrac{1}{n^{1/2}}$
Here, $\Sigma_{n=1}^\infty \dfrac{1}{n^{1/2}}$ a divergent p-series with $p=\dfrac{1}{2} \lt 1$
Hence, the series diverges by the comparison test.