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.
 A: 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:
$$\lim\limits_{N \to\infty} \frac{1}{\sqrt N}\sum_{n=1}^N\frac{1}{\sqrt n}=\int_0^1  \frac {1}{\sqrt x}dx=2$$
A: 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
$1.858960382478$
A: Firstly note that
$$2(\sqrt{n+1}-\sqrt{n}) = 2\frac{(\sqrt{n+1}-\sqrt{n}) (\sqrt{n+1}+\sqrt{n}) }{(\sqrt{n+1}+\sqrt{n}) }=\frac{2}{(\sqrt{n+1}+\sqrt{n}) }<\frac{2}{2\sqrt{n}}=\frac{1}{\sqrt{n}}$$
Similarly one can show that :
$$2(\sqrt{n}-\sqrt{n-1}) >\frac{1}{\sqrt{n}}$$
Clubbing the inequalities we have
$$2{(\sqrt{n+1}-\sqrt{n}) }<\frac{1}{\sqrt{n}}<2(\sqrt{n}-\sqrt{n-1}) $$
Summing them up from $n=1$ to $n=N$ we get a telescopic sum and hence the inequality becomes :
$$ 2(\sqrt{N+1}-1)<\sum \limits_{n=1}^{n=N} \frac{1}{\sqrt{n}}<2(\sqrt{N})$$
Now dividing by $\sqrt{N}$ yields
$$\frac{2(\sqrt{N+1}-1)}{\sqrt{N}}<\frac{1}{\sqrt{N}}\sum \limits_{n=1}^{n=N} \frac{1}{\sqrt{n}}<2$$
Letting $N\to \infty $ and using sandwich theorem ,we get the required limit equal to 2. That is :
$$\lim_{N\to \infty} \frac{1}{\sqrt{N}}\sum \limits_{n=1}^{n=N} \frac{1}{\sqrt{n}} =2$$
A: Let's use Cesàro-Stolz:
$\frac{\sum_{k=1}^{n+1} \frac{1}{\sqrt{k}} - \sum_{k=1}^{n} \frac{1}{\sqrt{k}}}{\sqrt{n+1} - \sqrt{n}} = \frac{1/\sqrt{n+1}}{\sqrt{n+1} - \sqrt{n}} = \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}}$ and this obviously converges to $2$ as $n \to \infty$.
