Show that this difference goes to zero,

Question

$$\frac{1+\sqrt{2} + ... + \sqrt{N}}{N} - \frac{2}{3}\sqrt{N} \to 0.$$

The hint given in the question is this: choose appropriate Riemann sums and estimate the approximation error.

My current work:

$$\frac{1+\sqrt{2} + ... + \sqrt{N}}{N} - \frac{2}{3}\sqrt{N}$$

$$=: A_n =(\sum_{k=1}^N \sqrt{k}\frac{1}{N}) - \frac{2}{3}\sqrt{N}$$

The first term is in the form of a Riemann sum, so letting N go to infinity, we see that mesh(p) goes to zero, for some partition p, which gives the (improper) Riemann integral, over the interval [1,N]:

$$\lim_{N->\infty}\int_1^N \sqrt{x}dx$$

Evaluation of the integral, without evaluating the limit, gives:

$$\frac{2}{3}N^{\frac{3}{2}} - \frac{2}{3}$$

Then $$A_n = \frac{2}{3}N^{\frac{3}{2}} - \frac{2}{3} - \frac{2}{3}\sqrt{N}$$

And this is where I am currently stuck. The above equation is a little suspect, because I let N go to infinity to get the improper integral, while I did nothing with the $\frac{2}{3}\sqrt{N}$ term -- and just included this term into the equation, since I feel it gets me a little closer to do some kind of approximation.

Any hints would be greatly appreciated.

Thanks,

Answer 1

I believe the proper Riemann Sum is $$ \begin{align} \frac1n\sum_{k=1}^nk^{1/2} &=n^{1/2}\sum_{k=1}^n\color{#C00000}{\left(\frac kn\right)^{1/2}}\,\color{#00A000}{\frac1n}\\ &=n^{1/2}\int_0^1\color{#C00000}{x^{1/2}}\,\color{#00A000}{\mathrm{d}x}+O\left(n^{-1/2}\right)\\[3pt] &=\frac23n^{1/2}+O\left(n^{-1/2}\right) \end{align} $$ since the error estimate for the Riemann Sum is $$ \begin{align} \frac1n\int_0^1\left|\,f'(x)\right|\,\mathrm{d}x &=\frac1n\int_0^12x^{-1/2}\,\mathrm{d}x\\ &=\frac1n \end{align} $$

Answer 2

If you partition the interval $[0,N]$ to $N$ equal length subintervals, then those subintervals all have length $1$. Not $1/N$ as you seem to think.

I think that you are expected to observe that the sum $$ \frac1{\sqrt{N}}\,\frac{\sqrt1+\sqrt2+\cdots+\sqrt{N}}N $$ is a Riemann sum related to the definite integral $$ \int_0^1\sqrt x\,dx. $$

In other words, scale everything by a factor of $\sqrt{N}$.

Standard error estimate available for all increasing functions is good enough unless I made a mistake.

Answer 3

We can also do this with Stolz-Cesaro. Let $S_n = \sum_{k=1}^{n}k^{1/2}.$ We are looking at

$$\frac{S_n -(2/3)n^{3/2}}{n}.$$

S-C says to look at

$$(n+1)^{1/2}+ (2/3)[n^{3/2} - (n+1)^{3/2}].$$

By the MVT, the second term equals $-c_n^{1/2},$ where $c_n \in (n,n+1).$ So we have

$$(n+1)^{1/2}-c_n^{1/2} = \frac{n+1-c_n}{(n+1)^{1/2}+c_n^{1/2}},$$

which is positive and less than $1/ (n+1)^{1/2} \to 0.$