Evaluate the limit $\lim_{n\to \infty }\sum_{i=1}^n \frac{1}{n} \cdot \lfloor \sqrt {\frac{4i}{n}} \rfloor$

Question

I solved the problem using the Riemann integral. However, my answer did not match with the result given in the book. My answer was $\frac{3}{4}$ and the answer given in the book was just 3.

Help me understand where I went wrong. My solution

$$\lim_{n\to\infty} \sum_{i=1}^n \frac{1}{n}\cdot \lfloor \sqrt{\frac{4i}{n}}\rfloor =\lim_{n\to\infty} \sum_{i=1}^n \frac{1}{n} \cdot \lfloor 2\sqrt{\frac{i}{n}}\rfloor=$$ $$=\lim_{n\to\infty}\left(\frac{1}{n}\cdot \lfloor2\sqrt{\frac{1}{n}}\rfloor+\frac{1}{n}\cdot \lfloor2\sqrt{\frac{2}{n}}\rfloor+\ldots +\frac{1}{n}\cdot \lfloor2\sqrt{\frac{n}{n}}\rfloor\right)$$

Clearly, the given expression is a Riemann sum of the function $\lfloor2\sqrt{x}\rfloor$ on the interval $[0,1]$.

$$\lim_{n\to\infty} \sum_{i=1}^n \frac{1}{n}\cdot \lfloor \sqrt{\frac{4i}{n}}\rfloor =$$

$$=\int_0^1 \lfloor2\sqrt{x}\rfloor dx=\int_0^{\frac14} \lfloor2\sqrt{x}\rfloor dx+\int_{\frac14}^1 \lfloor2\sqrt{x}\rfloor dx=0+1\cdot\left(1-\frac14\right)=\frac34$$

Answer 1

Your solution seems fine, to check we can also solve it in a direct way for $n=4N$

$$\sum_{i=1}^n \frac{1}{n}\cdot\lfloor\sqrt{\frac{4i}{n}}\rfloor=\sum_{i=1}^{4N} \frac{1}{4N}\cdot\lfloor\sqrt{\frac{i}{N}}\rfloor=$$

$$=\sum_{i=1}^{N-1} \frac{1}{4N}\cdot0+\sum_{i=N}^{4N-1} \frac{1}{4N}\cdot 1+\frac1{2N}=\frac{3}{4}+\frac1{2N} \to \frac34$$

Answer 2

Distinguish the three cases I < n/4, n/4 <= I < n, and I = n, and the result is trivial. This has nothing to do with any integral or Riemann sum.

The square root is 0 <= root < 1 if I < n/4, 1 <= root < 2 if n/4 <= I < n, and 2 if I = n. Rounded down to an integer it is 1 for n/4 <= I < n and 2 if I = n.

The sum therefore is (n - floor(n/4)) / n + 2/n, which is 3/4 + 5/4n <= sun <= 3/4 + 2/n, with limit 3/4.