"Lebesgue" Sum

$\begin{aligned} S & = \lim_{n \to \infty} \sum_{i=1}^n \frac in \left(\sqrt{\frac in} - \sqrt{\frac{i-1}n} \right) \\ & = \lim_{n \to \infty} \left[ \frac 1n \sqrt{\frac 1n} - \frac 1n \sqrt{\frac 0n}+ \frac 2n \sqrt{\frac 2n} - \frac 2n \sqrt{\frac 1n} + \frac 3n \sqrt{\frac 3n} - \frac 3n \sqrt{\frac 2n}+ ... + \color{#3D99F6}{\frac nn \sqrt{\frac nn}} - \frac nn \sqrt{\frac {n-1}n} \right] \\ & = \lim_{n \to \infty} \left[ \color{#3D99F6}{1} - \frac 1n \sum_{i=1}^n \sqrt{\frac in} \right] \\ & = 1 - \color{#3D99F6}{\lim_{n \to \infty} \frac 1n \sum_{i=1}^n \sqrt{\frac in} \quad \quad \small \text{Applying Riemann sums}} \\ & = 1 - \color{#3D99F6}{\int_0^1 x^\frac 12 dx} = 1 - \frac 23 = \frac 13 \end{aligned}$

$\implies a+b = 1+3 = \boxed{4}$

This sum is actually approximating $\int_0^1 x^2\, dx$ - it partitions $0,1$ on the $y$ -axis into $n$ increments. At each point, it multiplies that value by an approximation for the length of the interval of $x$ -values that produce that $y$ -value. As $n$ goes to infinity, this approximation becomes increasingly accurate, so the value of the limit is the same as the value of the integral, which is $1/3$ , so the answer is 4.

We have $\displaystyle \lim_{n\to\infty} \sum_{i=1}^{n}\frac{i}{n}\left(\sqrt{\frac{i}{n}}-\sqrt{\frac{i-1}{n}}\right)$

Multiplying and dividing by $\displaystyle \left(\sqrt{\frac{i}{n}}+\sqrt{\frac{i-1}{n}}\right)$ we have:-

$\displaystyle \lim_{n\to\infty} \sum_{i=1}^{n}\frac{\frac{i}{n}\frac{1}{n}}{\left(\sqrt{\frac{i}{n}}+\sqrt{\frac{i-1}{n}}\right)}$ $\displaystyle = \lim_{n\to\infty} \frac{1}{n}\sum_{i=1}^{n}\frac{\frac{i}{n}}{\left(\sqrt{\frac{i}{n}}+\sqrt{\frac{i-1}{n}}\right)}$

Using Riemann Sums we have :-

$\displaystyle \int_{0}^{1} \frac{x}{2\sqrt{x}}\,dx = \frac{1}{3}$ which is our answer.