Evaluate $\displaystyle\lim_{n\to\infty} \frac{a_n}{n^2}$ Problem :
Let $f\colon[0,\infty)\to\mathbb{R}\quad f(x)=\sin (\pi x)$.
For each $n\in\mathbb{N}$, let $x_1,x_2,\cdots,x_k$ denotes roots of $$ f(x)=\frac{x}{2n}$$
Evaluate
$$\lim_{n\to\infty} \frac{a_n}{n^2}$$
where $a_n$ is sum of all $x_1, x_2, \cdots, x_k$.

My Attempt
I think $\frac{a_n}{n^2}\to 2$.
From observation, the equation $$f(x)=\frac{x}{2n}$$ has $2n$ distinct real roots for each $n\in\mathbb{N}$.
Rewrite this roots as $x_1, x_2, \cdots, x_{2n}$.
If $n\to\infty$,  $\frac{x}{2n}\to 0$.
This means the equation will be changed to : $f(x)=0$ under $n\to\infty$.
And, real roots of $f(x)=0$ is well known : $0, 1, 2, \cdots$
So I think $$\left\{x_1,x_2,\cdots,x_{2n}\right\}\to\left\{0,1,\cdots,2n-1\right\}.$$
This means$$a_n \approx \sum_{k=0}^{2n-1}k = n(2n-1)$$
And finally $$\lim_{n\to\infty} \frac{a_n}{n^2} \approx \lim_{n\to\infty}\frac{n(2n-1)}{n^2}=2$$

I have two questions :

*

*Is this valid?


*Is there any nice approach like squeeze theorem?
(I think it's hard to represent $x_n$ with $\arcsin$, so I think squeeze theorem will be the best approach.)
 A: Let $x_{mn}$ be the $m^{th}$ root of $\sin\pi x=x/2n$ counting from zero, so $m$ ranges from $0,...,2n-1$ as noted. Then $a_n=\sum_{m=0}^{2n-1}x_{mn}$. As you said $\lim_{n\rightarrow \infty}x_{mn}=m$.
I will make a squeeze argument using some bounds for $x_{mn}$. Evaluate $\sin\pi x-\frac{x}{2n}$ at $m+\tfrac{1}{2}$ and we get alternating positive and negative answers, showing that there is a root in each interval $(m-\tfrac{1}{2},m+\tfrac{1}{2})$ for all $m=0,...,2n-1$. To get the bound, we also need to show there is at most one root in each interval, which I haven't figured out yet, but shouldn't be too hard. This shows that
$$m-\tfrac{1}{2}\le x_{mn}\le m+\tfrac{1}{2}$$
and hence
$$\frac{1}{n^2}\sum_{m=0}^{2n-1}(m-\tfrac{1}{2})\le\frac{1}{n^2}\sum_{m=0}^{2n-1}x_{mn}\le\frac{1}{n^2}\sum_{m=0}^{2n-1}(m+\tfrac{1}{2})$$
But the LHS and RHS both tend to $2$ as $n\rightarrow \infty$.
A: Write your equation as $\frac{\sin(\pi x)}{x} = \frac{1}{n}.$ It is clear that any solution is at most $n$, since if $x > n,$ the LHS is smaller than $1/n.$ Thus, the limit has to be zero.
