Series-Convergence and divergence problem 
Does this series converge or diverge?
  $$\sum_{n=2}^\infty\frac{n^{1/2}}{n-1}$$

I think it is divergent. This is how I proceeded: Divide numerator and denominator by $n^{1/2}$ and it is greater than or equal to $\frac1{n^{1/2}}$  and since this is divergent, by comparison test, given series is divergent. Am I done?
 A: Hint: You can use the comparison test and the fact that $$\frac{n^{1/2}}{n-1} > \frac{1}{n-1} > \frac1n$$
In your question, you used the comparison test with $\frac1{\sqrt{n}}$. This also works, but I would write out a little more specifically why the terms are larger than $\frac1{\sqrt{n}}$, like I did above with $\frac1n$. 
A: Limit Comparison Test + Integral Test:
$$\sum_{n=2}^\infty\frac{n^{1/2}}{n-1}\implies\sum_{n=2}^\infty\frac{n^{1/2}}n=\sum_{n=2}^\infty\frac1{n^{1/2}}\implies\int_2^\infty x^{-1/2}dx$$
A: We can see that by D'Alembert's ratio test the limit becomes 1 and thus the test fails. Then we use Raabe's test and after some manipulation we get the limit as 0.5 when n tends to infinity. Thus the series is divergent.
 And also the method you have used is perfectly right as the terms of the series are less than n^0.5 for all n>1 .
A: Simplest (as often): Use equivalents, since it is a series with positive terms. The general term is
$$\frac{n^{1/2}}{n-1}\sim_\infty\frac{n^{1/2}}{n}=\frac1{n^{1/2}},$$
which is a divergent Riemann series.
