Answer
Diverges, by Th. 9.12.
Work Step by Step
Th. 9.12 Direct Comparison Test:
Let $0 < a_{n} \leq b_{n}$ for all $n$.
1. If $\displaystyle \sum_{n=1}^{\infty}b_{n}$ converges, then $\displaystyle \sum_{n=1}^{\infty}a_{n}$ converges.
2. If $\displaystyle \sum_{n=1}^{\infty}a_{n}$ diverges, then $\displaystyle \sum_{n=1}^{\infty}b_{n}$ diverges.
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Observe that
$\displaystyle \frac{1}{\sqrt{n}-1} > \frac{1}{\sqrt{n}}$ for $n \geq 2.$
$0 < \displaystyle \frac{1}{\sqrt{n}} \leq \frac{1}{\sqrt{n}-1}$
$( 0 < a_{n} \leq b_{n} )$
(2).Since $\qquad \displaystyle \sum_{n=1}^{\infty}\frac{1}{\sqrt{n}} =\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^{1/2}}\quad $diverges,
($p=\displaystyle \frac{1}{2},$ divergent p-series, Th.9.11)),
so $\displaystyle \sum_{n=1}^{\infty}\frac{1}{\sqrt{n}-1}$ diverges.
