# What is the limit of $\left(2\sqrt{n}\left(\sqrt{n+1}-\sqrt{n}\right)\right)^n$ as $n \to \infty$?

I'd would like to know how to get the answer of the following problem:

$$\lim_{n \to \infty} \left(2\sqrt{n}\left(\sqrt{n+1}-\sqrt{n}\right)\right)^n$$

I know that the answer is $\frac{1}{e^{1/4}}$, but I can't figure out how to get there. This is a homework for my analysis class, but I can't solve it with any of the tricks we learned there.

This is what I got after a few steps, however it feels like this is a dead end:

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

• I suppose you meant "...as $\,n\to\infty\,$" , right? – DonAntonio Oct 7 '12 at 13:12
• It could be simplified to $\frac{2^n}{\big(1+\sqrt{1+\frac{1}{n}}\big)^n}$ as well. – Nancy Rutkowskie Oct 7 '12 at 13:22
HINT: Use $$\sqrt{n+1}-\sqrt{n} = \frac{\left(\sqrt{n+1}-\sqrt{n}\right) \left(\sqrt{n+1}=\sqrt{n}\right) }{\left(\sqrt{n+1}+\sqrt{n}\right) } = \frac{1}{\sqrt{n+1}+\sqrt{n}}$$ Now you limit becomes easier to handle: $$\lim_{n \to \infty} \left(\frac{2 \sqrt{n}}{\sqrt{n+1} + \sqrt{n} }\right)^n = \lim_{n \to \infty} \left(1 - \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n+1} + \sqrt{n} }\right)^n = \lim_{n \to \infty} \left(1 - \frac{1}{\left(\sqrt{n+1} + \sqrt{n}\right)^2 }\right)^n$$
Expanding the Taylor series of $\sqrt{1+x}$ near $x=0$ gives $\sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \mathcal{O}(x^3).$ Thus $$\left( 2\sqrt{n} ( \sqrt{n+1} - \sqrt{n} )\right)^n =2^n n^{\frac{n+1}{2}} \left(\sqrt{1+\frac{1}{n}}-1 \right)^n$$
$$=2^n n^{\frac{n+1}{2}} \left( \frac{1}{2n} - \frac{1}{8n^2}+ \mathcal{O}(1/n^3)\right)^n = n^{\frac{1-n}{2}} \left(1 - \frac{1}{4n} + \mathcal{O}(n^{-2})\right)^n\to e^{-1/4}.$$