What is the value of $\lim\limits_{n\to \infty}\left(\sum\limits_{i=1}^n{ \frac{1}{\sqrt{i \cdot n}} } \right)$? [duplicate]

How can I calculate this limit: $$\lim_{n\to \infty} \Bigg(\sum_{i=1}^n{ \frac{1}{\sqrt{ \vphantom{1}i \cdot n}} } \Bigg)$$

Note: I think it has something to do with some Riemann sum, but I can't get anywhere, I tried to set $n$ as the number summands and $\frac1n$ as the width of each part. But that doesn't help; it gives: $\frac1n\sum\frac{\sqrt{n}}{\sqrt{i}}$

Solution 1:

Although the sum at hand can be rewritten as a lower Riemann sum for an integral $\displaystyle\;\int_0^1 \frac{dx}{\sqrt{x}}$. $$\sum_{k=1}^n \frac{1}{\sqrt{kn}} = \frac1n\sum_{k=1}^n \sqrt{\frac{n}{k}}$$ It doesn't mean we can directly claim the limit of the sum equal to the integral. This is because the intergrand $\frac{1}{\sqrt{x}}$ diverges at $x = 0$ and the upper Riemann sum of the integral doesn't exist. To proceed along this route, more justification is needed.

If you really want to go along this route, look at this answer for a related question. It shows that when your intergrand is non-negative, decreasing and the improper Riemann integral exists, the limit of the sum still equal to the improper Riemann integral.

In this answer, we use a different approach, we will bound the sum directly.

For $k \ge 1$, we have $\displaystyle\;\frac{\sqrt{k}+\sqrt{k-1}}{2} \le \sqrt{k} \le \frac{\sqrt{k+1}+\sqrt{k}}{2}$, this leads to $$2(\sqrt{k}-\sqrt{k-1}) = \frac{2}{\sqrt{k}+\sqrt{k-1}} \ge \frac{1}{\sqrt{k}} \ge \frac{2}{\sqrt{k+1}+\sqrt{k}} = 2(\sqrt{k+1}-\sqrt{k}) $$ Summing $k$ from $1$ to $n$ and take advantage of the fact that the terms on LHS and RHS are telescoping, we obtain

$$ 2\sqrt{n} \ge \sum_{k=1}^n \frac{1}{\sqrt{k}} = 2(\sqrt{n+1} - 1) \ge 2\sqrt{n} - 2 \quad\implies\quad 2 \ge \sum_{k=1}^n \frac{1}{\sqrt{nk}} \ge 2 - \frac{2}{\sqrt{n}} $$ Sim $\lim\limits_{n\to\infty} 2 - \frac{2}{\sqrt{n}} = 2$, by squeezing, we can conclude $$\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{\sqrt{kn}} = 2$$

Solution 2:

You're on the right track. Just think of ${1\over n}\sum{\sqrt n\over\sqrt i}$ as ${1\over n}\sum f(i/n)$ with $f(x)={1\over\sqrt x}$. Can you take it from there?