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.
Best Answer
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:
$$2-\epsilon -\epsilon <2\sqrt{1+a}-2\sqrt a-\epsilon<\frac{1}{N}\sum_{n=1}^N f (\frac{n}{N})=\frac{1}{\sqrt N}\sum_{n=1}^N\frac{1}{\sqrt n}<2.$$