# 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?

• What do you think ? What did you try already ? – Claude Leibovici Sep 25 '16 at 10:32
• I think its divergent. This is how i proceeded divide numerator and denominator by n^1/2 and it is greater than or equal to 1/(n)^1/2 and since this is divergent..by comparison test, given series is divergent...am I done? – user371841 Sep 25 '16 at 10:33
• You can use the limit comparison test, for instance. – Mark Sep 25 '16 at 10:34
• Please post your work. Otherwise, there are big chances that the question will be closed for missing context. – Claude Leibovici Sep 25 '16 at 10:34
• Better ! And, by the way, welcome to MSE which is a fantastic site. Cheers :-) – Claude Leibovici Sep 25 '16 at 10:39

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 $\frac1n$.
$$\sum_{n=2}^\infty\frac{n^{1/2}}{n-1}\implies\sum_{n=2}^\infty\frac{n^{1/2}}n=\sum_{n=2}^\infty\frac1{n^{1/2}}\implies\int_2^\infty x^{-1/2}dx$$
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.