<s>[INST] Why $\lim\limits_{n \to \infty} \frac{\frac1{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim\limits_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}}$? I just need clarification about answer to this question:

$$\lim_{n \to \infty} \frac{\sum\limits_{i = 1}^{n} \frac{1}{\sqrt{i}}}{\sqrt{n}} = \lim_{n \to \infty} \frac{\sum\limits_{i = 1}^{n+1} \frac{1}{\sqrt{i}} - \sum\limits_{i = 1}^{n} \frac{1}{\sqrt{i}}}{\sqrt{n+1} - \sqrt{n}} = \lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}} = 2$$
$$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n}} \sum\limits_{i = 1}^{n} \frac{1}{\sqrt{i}}\right)^k = 2^k$$


In this case I don't understand why
$$\lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}}$$
My reasoning. $$\frac{ \frac{1}{a} }{b - c} = \frac{1}{a(b-c)}$$Hence, we get:
$$\lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}}=\lim_{n \to \infty} \frac{1}{n+1 - \sqrt{n}\sqrt{n+1}} = \lim_{n \to \infty} \frac{ \frac{1}{n} }{ \frac{n}{n} + \frac{1}{n} + \sqrt{\frac{n^2}{n^2}+\frac{n}{n^2} } } = \lim_{n \to \infty}\frac{0}{1+0+\sqrt{1+0}}=0$$
While $\lim_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}} = \lim_{n \to \infty}\frac{ \sqrt{\frac{n}{n} + \frac{1}{n}} + \sqrt{\frac{n}{n}} }{ \sqrt{\frac{n}{n} + \frac{1}{n} } } = \lim_{n \to \infty}\frac{\sqrt{1+0} + \sqrt{1} }{\sqrt{1+0}} = 2$
Any explanation, what's going on? Thank you
 [/INST] Let $s = \sqrt{n+1}$ and $t = \sqrt{n}.$ Then:
$$\mathcal{L} = \lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim_{n \to \infty} \frac{\frac{1}{s}}{s-t}  = \\
= \lim_{n \to \infty}\displaystyle\frac{1}{(s-t)}\frac{1}{s} = \lim_{n \to \infty}\displaystyle\frac{s+t}{(s-t)(s+t)}\frac{1}{s}.$$
But:
$$(s-t)(s+t) = s^2 -t^2 = (n+1) - n = 1,$$
and hence:
$$\mathcal{L} = \lim_{n \to \infty} \frac{s+t}{1}\frac{1}{s} = \lim_{n \to \infty} \frac{s+t}{s}= \lim_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}}.$$
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