Finding $\lim_{n \rightarrow +\infty}{\frac{\sum_{i=1}^{n} \lfloor{i \sqrt{2}}\rfloor}{n^2}}$ Find this limit:
$\lim_{n \rightarrow +\infty}{\frac{\sum_{i=1}^{n} \lfloor{i \sqrt{2}}\rfloor}{n^2}}$
It is like that we want to find the limit of the sequence $\frac{[\sqrt2] +[2\sqrt2] + ... + [n\sqrt2]}{n^2} $ at infinity.
At the first glance, It seem it should be like $0$ but I somewhere read that its limit is $\frac{\sqrt2}{2}$ or something like that. I know that it must be something other than zero but is it right that its limit is $\frac{\sqrt2}{2}$? If yes how? and if it is not, then what is its limit at infinity?
 A: For every $i$, you have by definition of the floor function
$$
i\sqrt{2} - 1\leq \lfloor i\sqrt{2}\rfloor\leq i\sqrt{2} \tag{1}
$$
From this, summing for $i$ ranging from $1$ to $n$ and dividing by $n^2$,
$$
\sqrt{2}\sum_{i=1}^n \frac{i}{n^2} - \frac{n}{n^2} 
\leq \sum_{i=1}^n \frac{\lfloor i\sqrt{2}\rfloor}{n^2}
\leq \sqrt{2}\sum_{i=1}^n \frac{i}{n^2} \tag{2}
$$
and recalling that $\sum_{i=1}^n i = \frac{n(n+1)}{2}$, we get
$$
\sqrt{2}\frac{(n+1)}{2n} - \frac{1}{n} 
\leq \sum_{i=1}^n \frac{\lfloor i\sqrt{2}\rfloor}{n^2}
\leq \sqrt{2}\frac{(n+1)}{2n}\,. \tag{3}
$$
By the squeeze theorem, the limit is therefore
$$
\boxed{\lim_{n\to\infty }\sum_{i=1}^n \frac{\lfloor i\sqrt{2}\rfloor}{n^2} = \frac{\sqrt{2}}{2}\,.}
$$
A: If you do not have the floor function, then the result follows easily from the identity $\sum_{i=1}^n i = n(n-1)/2$.
$$\frac{\sum_{i=1}^n (i \sqrt{2})}{n^2} = \sqrt{2} \frac{\frac{n(n-1)}{2}}{n^2} \to \frac{\sqrt{2}}{2}.$$

The idea is that the floor function does not really change much, and will have the same behavior in the limit.
To show this rigorously, find two functions $f$ and $g$ such that
$$f(n) \le \frac{\sum_{i=1}^n \lfloor i \sqrt{2} \rfloor}{n^2} \le g(n)$$
for all $n$, and such that $f$ and $g$ both tend to $\sqrt{2}/2$ in the limit.
Hint: I already found $g$ for you.

 Take $g=\frac{\sum_{i=1}^n (i \sqrt{2}) }{n^2}$ and $f = \frac{\sum_{i=1}^n (i\sqrt{2} - 1)}{n^2}$.

A: Since $\sqrt{2}\not\in\mathbb{Q}$, the fractional part $\{k\sqrt{2}\}$ is equidistributed $\!\!\pmod{1}$. In particular
$$ \sum_{k=1}^{n}\lfloor k\sqrt{2}\rfloor = \sqrt{2}\,\frac{n(n+1)}{2}-\sum_{k=1}^{n}\{k\sqrt{2}\}=\frac{1}{\sqrt{2}}n(n+1)-\frac{n}{2}+o(n)$$
and we even know the second term of the asymptotic expansion:
$$ \frac{1}{n^2}\sum_{k=1}^{n}\lfloor k\sqrt{2}\rfloor = \frac{1}{\sqrt{2}}+\color{red}{\frac{1}{(2+2\sqrt{2})n}}+o\left(\frac{1}{n}\right).$$
