Let \(a_n =\sqrt{n}\left(\sqrt{n+1}-\sqrt{n}\right)\). Prove that \(\lim_{n\to\infty}a_n=1/2\).
Solution
1. \(a_n = \sqrt{n}(\sqrt{n+1} - \sqrt{n}) = \sqrt{n}\cdot\frac{(n+1) - n}{\sqrt{n+1} + \sqrt{n}} = \frac{\sqrt{n}}{\sqrt{n+1} + \sqrt{n}}\).
rationalize: \(\sqrt{n+1} - \sqrt{n} = \frac{(n+1)-n}{\sqrt{n+1}+\sqrt{n}}\).
2. \(a_n = \dfrac{1}{\frac{\sqrt{n+1}}{\sqrt{n}} + 1} = \dfrac{1}{\sqrt{1 + 1/n} + 1}\).
divide numerator and denominator by \(\sqrt{n}\).
3. \(\sqrt{1 + 1/n} \to 1\) as \(n \to \infty\).
continuity of the square root and \(1 + 1/n \to 1\) (or the standard squeeze \(1 \le \sqrt{1+1/n} \le 1 + 1/(2n)\)-style bound).
4. Q.E.D.: \(a_n \to \dfrac{1}{1 + 1} = \dfrac{1}{2}\).
2 and 3, by the quotient law for limits.