Find a limit with Gauss function

Question

How can I find the following limit:

$$\lim_{n\to\infty}\frac{1}{n^2}\sum_{i,j=1}^n [\frac{4i+9j}{n}]$$ where $[x]$ is the maximal integer less than $x$, for example, $[3]=3$, $[3.5]=3$.

Answer 1

I put $A=\{(i,j); 1\leq i\leq n; 1\leq j\leq n\}$.

a) Let first be $x\in \mathbb{Q}$ such that $x\not \in \mathbb{N}$ and $0\leq x\leq 13$. Then we have $[x[<x<[x]+1$, and $12-[x]<13-x<13-[x]$; it follows that $[13-x]=12- [x]$.

b) Let $B=\{(i,j); 1\leq i\leq n-1; 1\leq j\leq n-1; (4i+9j)/n\not \in \mathbb{N}\}$

Then the application $(i,j)\to (n-i,n-j)$ is a bijection of $B$ on $B$.(As $B$ is finite, it suffices to show that this is an injective application of $B$ to $B$).

c) Let $C=A\setminus{B}$. Then if $(i,j)\in C$, with $i\not =n$ and $j\not =n$, there exists $k$ with $1\leq k\leq 13$ such that $4i+9j=kn$. For a fixed $k$, for each $i$, $1\leq i\leq n-1$, there is at most one $j$ such that $4i+9j=kn$. Hence the cardinal of $C$ is bounded by $13n$+$2n$ =$15n$ (the additional term is for the couples $(n,j)$ and $(i,n)$). Hence $$n^2-15n\leq {\rm Card}( B)\leq n^2$$ and we have also

$$U_n=\sum_{(i,j)\in C}[\frac{4i+9j}{n}]\leq 13{\rm Card}(C)\leq 13\times 15 n=c n$$

d) Let $\displaystyle T_n=\sum_{(i,j)\in B}[\frac{4i+9j}{n}]$. We have, using b):

$$T_n=\sum_{(i,j)\in B}[\frac{4(n-i)+9(n-j)}{n}]=12{\rm Card}(B)-T_n$$ hence $T_n=6 {\rm Card}(B)$.

e) Let $\displaystyle S_n=\sum_{(i,j)\in A}[\frac{4i+9j}{n}]$. We have $S_n=T_n+U_n$, and hence $$6(n^2-15n)\leq S_n\leq 6n^2+cn$$ Hence $\displaystyle \frac{S_n}{n^2}\to 6$ as $n\to \infty$.