Proving Convergence or Divergence using Comparison test

I'm asked to prove the convergence or divergence of the below series using the comparison test.

$$\sum_{n=1}^{\infty}\sqrt{(n-1)}-\sqrt{n}$$

I've reduced the above series to $\sqrt{n}(\sqrt{(1-\frac{1}{n}})-1)=U_n$, but after taking $V_n$ as $\sqrt{n}$ and applying the limit, $\lim_{n\to {\infty}}\frac{U_n}{V_n}$ yields me zero, but the value has to be greater than zero for the comparison test to work. So I'm in a dilemma, am I doing something wrong?

Feel free to correct me if I'm wrong in my calculations.

• What's the meaning of the first and the second $=$ signs from your question? Sep 9 '18 at 16:18
• I've assigned the series to $U_n$ Sep 9 '18 at 16:21
• That doesn't answer my question. I want to know the meaning of the $=$ sign in $$\sum_{n=1}^{\infty}=\sqrt{(n-1)}-\sqrt{n},$$for instance. Sep 9 '18 at 16:22
• Sorry, I'm in a hurry I didn't notice that. Sep 9 '18 at 16:28

$$\sqrt{n}-\sqrt{(n-1)}\ge \frac13 \frac1{n^{2/3}}$$
Firstly, remember that $x^{3} - y^{3} = (x-y)(x^{2} + xy +y^{2})$. Hence we get: \begin{align*} \sqrt{n} - \sqrt{n-1} & = \sqrt{n} - \sqrt{n-1}\times\frac{\sqrt{n^{2}} + \sqrt{n(n-1)} + \sqrt{(n-1)^{2}}}{\sqrt{n^{2}} + \sqrt{n(n-1)} + \sqrt{(n-1)^{2}}}\\ & = \frac{1}{\sqrt{n^{2}} + \sqrt{n(n-1)} + \sqrt{(n-1)^{2}}} \geq \frac{1}{3\sqrt{n^{3}}} \end{align*}