Interesting Limit (12)

Here's a proof from first principles. First we apply the algebraic identity $\sqrt{x+a} - \sqrt{x+b} = \frac{(\sqrt{x+a} - \sqrt{x+b})(\sqrt{x+a} + \sqrt{x+b})}{\sqrt{x+a} + \sqrt{x+b}} = \frac{a-b}{\sqrt{x+a}+\sqrt{x+b}}$ several times to write $\begin{aligned} \sqrt{n+2} - 2\sqrt{n+1} + \sqrt{n} &= \left(\sqrt{n+2} - \sqrt{n+1}\right) - \left(\sqrt{n+1} - \sqrt{n}\right) \\ &= \frac{1}{\sqrt{n+2}+\sqrt{n+1}} - \frac{1}{\sqrt{n+1}+\sqrt{n}} \\ &= \frac{\sqrt{n} - \sqrt{n+2}}{(\sqrt{n+2}+\sqrt{n+1})(\sqrt{n+1}+\sqrt{n})} \\ &= \frac{-2}{(\sqrt{n+2}+\sqrt{n+1})(\sqrt{n+2}+\sqrt{n})(\sqrt{n+1}+\sqrt{n})} \end{aligned}$ .

Therefore $\begin{aligned} \lim_{n\to\infty} n^{3/2}\left(\sqrt{n+2} - 2\sqrt{n+1} + \sqrt{n}\right) &= \lim_{n\to\infty} \frac{-2n^{3/2}}{(\sqrt{n+2}+\sqrt{n+1})(\sqrt{n+2}+\sqrt{n})(\sqrt{n+1}+\sqrt{n})} \\[1.5ex] &= \lim_{n\to\infty} \frac{-2}{\left(\sqrt{1+\frac{2}{n}}+\sqrt{1+\frac{1}{n}}\right)\left(\sqrt{1+\frac{2}{n}}+1\right)\left(\sqrt{1+\frac{1}{n}}+1\right)} \\[1.5ex] &= \frac{-2}{(\sqrt{1+0}+\sqrt{1+0})(\sqrt{1+0}+1)(\sqrt{1+0}+1)} \\ &= \frac{-2}{2^3} = \boxed{-\frac{1}{4} = -0.25} \end{aligned}$

$\begin{aligned} L & = \lim_{n \to \infty} n^\frac 32 \left(\sqrt{n+2}-2\sqrt{n+1} + \sqrt n\right) \\ & = \lim_{n \to \infty} n^2 \left({\color{#3D99F6}\sqrt{1+\frac 2n}}-2{\color{#3D99F6}\sqrt{1+\frac 1n}} +1\right) \qquad \qquad \small \color{#3D99F6} \text{By Taylor series expansion} \\ & = \lim_{n \to \infty} n^2 \left({\color{#3D99F6}\left(1+\frac 1n - \frac 1{2n^2} + \frac 1{2n^3}-\cdots\right)}-2{\color{#3D99F6}\left(1+\frac 1{2n} - \frac 1{8n^2} + \frac 1{16n^3}-\cdots\right)} +1\right) \\ & = \lim_{n \to \infty} n^2 \left(-\frac 1{4n^2} + \frac {3}{8n^3} - \cdots\right) \\ & = - \frac 14 = \boxed{-0.25} \end{aligned}$