<s>[INST] Testing convergence of the series of $n^p((n-1)^{-1/2}-n^{-1/2})$ 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.
 [/INST] 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.
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