How do I find the following limit?
$$
\lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + ... + \sqrt{n}}{n\sqrt{n}}
$$
The answer (from Wolfram) is $\frac{2}{3}$, but I'm not sure how to proceed.
Is this an application of the Squeeze theorem? I'm not quite sure.
Answer
$$\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\sqrt{\frac{k}{n}} = \int_{0}^{1}\sqrt{x}\,dx = \color{red}{\frac{2}{3}}$$
by Riemann sums and the integrability of $\sqrt{x}$ over $[0,1]$.
For a more elementary approach, notice that $\sqrt{k}$ is pretty close to $\frac{2}{3}\left[\left(k+\frac{1}{2}\right)^{3/2}-\left(k-\frac{1}{2}\right)^{3/2}\right]$ and apply creative telescoping and squeezing.
No comments:
Post a Comment