convergence of series $ {n}^{p} \cdot \left\{ \frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right\}$ $$ {n}^{p} \cdot \left\{ \frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right\}$$
determine the convergence of the series
$$ {n}^{p} \cdot \left\{ \frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right\} < {n}^{p} \cdot \frac{1}{\sqrt{n-1}\sqrt{n}} < \frac{{n}^{p}}{n-1} < \frac{1}{{n}^{1-p}-{n}^{-p}} < \frac{1}{{n}^{1-p}} $$
so converges when p<0 how about other cases?
 A: $$\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}=\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}\sqrt{n-1}}=\frac{1}{\sqrt{n^2-n}(\sqrt{n}+\sqrt{n-1})}=\frac{1}{n^{3/2}}\frac{1}{\sqrt{1-\frac{1}{n}}\left(1+\sqrt{1-\frac{1}{n}}\right)}\approx \frac{1}{2n^{3/2}}$$
So use the limit comparison test between $n^p\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right)$ and $n^{p-3/2}$.
A: HINT:
Using the Generalized Binomial Theorem, we can write
$$\begin{align}
\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt n}&=(n-1)^{-1/2}-n^{-1/2}\\\\
&=n^{-1/2}\left(1+\frac{1}{2n}+O\left(\frac1{n^2}\right)-1\right)\\\\
&=\frac12n^{-3/2}+O\left(\frac1{n^{5/2}}\right)
\end{align}$$
A: Hint:
$$\begin{align}
\frac{1}{\sqrt{n-1}} - \frac{1}{\sqrt{n}}
&= \frac{1}{\sqrt{n}}\left(  \frac{1}{\sqrt{1-\frac{1}{n}}} - 1\right)
= \frac{1}{\sqrt{n}}\left(  1+\frac{1}{2n} + o\left(\frac{1}{n}\right) - 1\right) \\
&= \frac{1}{2n^{3/2}} + o\left(\frac{1}{n^{3/2}}\right)
\end{align}$$
using Taylor expansions: $(1+x)^\alpha = 1+\alpha x + o(x)$ for any fixed $\alpha \in \mathbb{R}$, when $x\to 0$.
