A quite Riemann sum with square root

Question

I would like to compute the following limit:

$$\lim_{N\to\infty}\frac{1}{N^2}\sum_{i,j=0}^{N-1} \sqrt{i+1}\sqrt{j+1}.$$

It's looks like a Riemann sum minus a factor $\frac{1}{N}$ is the square root. Heuristically the limit would be $$\int_0^1 \int_0^1 \sqrt{x}\sqrt{y}dxdy$$.

I am not sure how to proceed.

Answer 1

$$f(n)=\frac 1{n^2}\sum_{i,j=0}^{n-1} \sqrt{i+1}\sqrt{j+1}=\frac 1{n^2}\left(\zeta \left(-\frac{1}{2},n+1\right)-\zeta \left(-\frac{1}{2}\right)\right)^2$$

Expanded as a series for large $n$ $$f(n)=\frac{4 }{9}n+\frac{2}{3}+\frac{4 \zeta \left(-\frac{1}{2}\right)}{3 \sqrt{n}}\left(1+A \right)$$ where $$A=\frac 1{\sqrt n}\left(\frac{11}{48 \zeta \left(-\frac{1}{2}\right)}+\frac 3{4 \sqrt n}+O\left(\frac{1}{n}\right)\right)$$ which is extreely accurate even for small values of $n$. For $n=5$, the absolute diffeence is $0.003$ for a function value of $2.810$.

The slope at the origin is $$-\frac{1}{8} \zeta \left(\frac{1}{2}\right) \zeta \left(\frac{3}{2}\right) \sim 0.476874$$ and at, for very large values of $n$ it tends, to $\frac 4 9$

Answer 2

$$\lim_{N\to\infty}\frac{1}{N^2}\sum_{i,j=0}^{N-1} \sqrt{i+1}\sqrt{j+1}=\lim_{N\to\infty}\left(\frac1N\sum_{i,j=0}^{N-1} \sqrt{i+1}\right)^2.$$

By the integral test, the limit diverges.