Testing convergence of the series of $n^p((n-1)^{-1/2}-n^{-1/2})$

Question

Exercise 8.15 (l) of Analysis by Apostol states:

Test for convergence: $$\sum\limits_{n = 2}^\infty n^p\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}} \right)$$

The solution I have states as the first step: "Note that $$ n^p\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}} \right) = \frac{1}{n^{3/2-p}}\left( \sqrt{\frac{n}{n-1}} \frac{1}{1+\sqrt{\frac{n-1}{n}}} \right)"$$

This step is not very self evident for me, could some one help me understand how we would obtain this equality.

Answer 1

You start with (ignoring the $n^p$ for the time being)

$$\frac{1}{\sqrt{n-1}} - \frac{1}{\sqrt{n}} = \frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n-1}\sqrt{n}},$$

then you rationalise the numerator, using $\sqrt{a}-\sqrt{b} = \frac{a-b}{\sqrt{a}+\sqrt{b}}$, getting

$$\frac{n-(n-1)}{\sqrt{n-1}\sqrt{n}(\sqrt{n}+\sqrt{n-1})}.$$

Now the numerator is $1$, and when you extract $(\sqrt{n})^3$ from the denominator, you get

$$\frac{1}{n^{3/2}}\frac{1}{\sqrt{\frac{n-1}{n}}\left(1 + \sqrt{\frac{n-1}{n}}\right)},$$

from which the given form is easily obtainable.