Question on evaluation of a limit of a sequence.

Find the following limit $:$

$$\lim_{N \to \infty} \frac {1} {\sqrt {N}} \sum\limits_{n=1}^{N} \frac {1} {\sqrt {n}}.$$

It is quite clear that $\sum\limits_{n=1}^{N} \frac {1} {\sqrt {n}} \gt \sqrt {N}$ for $N \gt 1.$ So the limit (if it exists finitely) has to be $\geq 1.$ But I believe that the limit is infinty. For that I need the sum to be greater than some scalar multiple of $N^s$ for sufficiently large $N$ where we require $s \gt \frac {1} {2}.$ Is it possible to attain this lower bound eventually? Any help in this regard would be greatly appreciated.

Thanks a lot.

I think it will helpful to those who are not familiar with the technique mentioned above in the comment section. So, I am posting an answer.

By definition 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:

$$\lim\limits_{N \to\infty} \frac{1}{\sqrt N}\sum_{n=1}^N\frac{1}{\sqrt n}=\int_0^1 \frac {1}{\sqrt x}dx=2$$

If you enjoy generalized harmonic numbers, you can have a good appromation os the partial sum $$S_N=\frac {1} {\sqrt {N}} \sum\limits_{n=1}^{N} \frac {1} {\sqrt {n}}=\frac {1} {\sqrt {N}}\,H_N^{\left(\frac{1}{2}\right)}$$ and, using asymptotics $$S_N=2+\frac{\zeta \left(\frac{1}{2}\right)}{\sqrt{N}}+\frac{1}{2 N}-\frac{1}{24 N^2}+O\left(\frac{1}{N^4}\right)$$ Using this trucated series for $N=100$, you would obtain $1.858960382452$ while the exact value is $1.858960382478$