I think it will be helpful to those who are not familiar with the technique mentioned above in the comment section. So, I am posting an answer.
By the definition of Riemann integrals, we know $\int_0^1 f(x) dx=\lim\limits_{N \to\infty} \frac{1}{N}\sum_{n=1}^N f (\frac{n}{N})$.
Also, $$\lim\limits_{N \to\infty} \frac{1}{\sqrt N}\sum_{n=1}^N\frac{1}{\sqrt n}=\lim\limits_{N \to\infty}\frac{1}{N} \sum_{n=1}^N \sqrt\frac{N}{n}=\lim\limits_{N \to\infty} \frac{1}{N}\sum_{n=1}^N f (\frac{n}{N}),$$
where $f(x)=\frac {1}{\sqrt x}$.
Therefore:
$$\lim\limits_{N \to\infty} \frac{1}{\sqrt N}\sum_{n=1}^N\frac{1}{\sqrt n}=\int_0^1 \frac {1}{\sqrt x}dx=2.$$
Update:
As mentioned in the comments, $f(x)=\frac{1}{\sqrt x}$ is not Riemann integrable on $[0,1]$, so some explanation is required to justify the last step. Indeed, we should be more careful with the approximation sum.
Note that $f(x)=\frac{1}{\sqrt x}$ is Riemann integrable on $[a,1+a]$ for $a>0$, and in particular, let's assume $2-\epsilon<2\sqrt{1+a}-2\sqrt a$, where $\epsilon$ is an arbitrary positive real number. Hence, by the definition of Riemann integrals, we have:
$$2\sqrt{1+a}-2\sqrt a=\int_a^{1+a} \frac{1}{\sqrt x} dx=\int_a^{1+a} f(x) dx=\lim\limits_{N \to\infty} \frac{1}{N}\sum_{n=1}^N f (a+\frac{n}{N}).$$
So, by the definition of limit, there is $N_0 \in \mathbb N$, such that for every $N>N_0$, we have:
$$2\sqrt{1+a}-2\sqrt a-\epsilon <\frac{1}{N}\sum_{n=1}^N f (a+\frac{n}{N})<2\sqrt{1+a}-2\sqrt a+\epsilon.$$
But for every $N$, we also have:
$$\frac{1}{N}\sum_{n=1}^N f (a+\frac{n}{N})<\frac{1}{N}\sum_{n=1}^N f (\frac{n}{N})=\frac{1}{\sqrt N}\sum_{n=1}^N\frac{1}{\sqrt n}<\frac{1}{\sqrt N}\sum_{n=1}^N 2(\sqrt n-\sqrt{n-1})=2.$$
So, for $N>N_0:$
$$2\sqrt{1+a}-2\sqrt a-\epsilon<\frac{1}{N}\sum_{n=1}^N f (\frac{n}{N})=\frac{1}{\sqrt N}\sum_{n=1}^N\frac{1}{\sqrt n}<2,$$
and:
$$2-\epsilon -\epsilon <2\sqrt{1+a}-2\sqrt a-\epsilon<\frac{1}{N}\sum_{n=1}^N f (\frac{n}{N})=\frac{1}{\sqrt N}\sum_{n=1}^N\frac{1}{\sqrt n}<2.$$
Best Answer
Using the binomial theorem, $$ \begin{align} \left(1+\sqrt{\frac2n}\right)^n &\ge1\color{#AAA}{+n\sqrt{\frac2n}}+\frac{n(n-1)}2\frac2n\color{#AAA}{+\cdots}\\ &\ge n\color{#AAA}{+\sqrt{2n}+\cdots} \end{align} $$ Therefore, $1\le n^{1/n}\le1+\sqrt{\frac2n}$. thus, $$ \lim_{n\to\infty}\left(n^{1/n}-1\right)=0 $$
Using the Taylor series for $e^x$, $$ \begin{align} n^{1/n} &=e^{\log(n)/n}\\ &=1+\frac{\log(n)}n+\frac12\left(\frac{\log(n)}n\right)^2+\frac16\left(\frac{\log(n)}n\right)^3+\cdots \end{align} $$ Therefore, $$ n^{1/n}-1\ge\frac{\log(n)}n $$
Since 1. is true, 3 cannot be.
Bernoulli's Inequality says $\left(1+\frac1{\sqrt{n}}\right)^{\sqrt{n}}\ge2$. Because $\frac{\log(x)}x$ is decreasing for $x\gt e$, for $n\ge16$, $\frac{\log\left(\sqrt{n}\right)}{\sqrt{n}}\le\frac{\log(2)}2$; that is, $n\le2^{\sqrt{n}}$. Thus, $$ \begin{align} n^{1/n} &\le2^{1/\sqrt{n}}\\ &\le1+\frac1{\sqrt{n}} \end{align} $$ Therefore, for $n\ge16$, $$ n^{1/n}-1\le\frac1{\sqrt{n}} $$