Evaluating Summation of Infinite Series

Let $$a_{1},a_{2},...,a_{n}$$ be a monotone increasing sequence of numbers satisfying $$\sin(a_{k}) = \frac{k}{n}$$ and $$a_{k} \leq \frac{\pi}{2}$$ for all $$1\leq k \leq n$$. I would like to calculate $$\lim\limits_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}a_{k}$$.

This is my attempt so far. First, fix $$n\in\mathbb{N}$$. Then, we have the upper bound as follows : $$\frac{1}{n}\sum_{k=1}^{n}a_{k} \leq \frac{\pi}{2}$$ Therefore, $$\lim\limits_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}a_{k}\leq \frac{\pi}{2}$$

On the other hand, I also have the following inequality: \begin{align*} \frac{1}{n}\sum_{k=1}^{n}\sin(a_{k})\leq \frac{1}{n}\sum_{k=1}^{n}a_{k} \end{align*} Moreover, we know that $$\sin(a_{k}) =\frac{k}{n}$$ and therefore $$\frac{1}{n}\sum_{k=1}^{n}\sin(a_{k})=\frac{n+1}{2n}$$ Hence, we have $$\frac{n+1}{2n}\leq \frac{1}{n}\sum_{k=1}^{n}a_{k}$$ which implies $$\frac{1}{2} \leq \lim\limits_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}a_{k}$$

Therefore, I obtain $$\frac{1}{2} \leq \lim\limits_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}a_{k} \leq \frac{\pi}{2}$$.

Now, I am confused how to find a better bound of this sum to apply the squeeze theorem. Any help or hint will be much appreciated! Thank you!

Hint. Let $$y=\sin(x)$$, then your sum is a Riemann sum with respect to the interval $$[0,1]$$ along the $$y$$-axis: $$\frac{1}{n}\sum_{k=1}^{n}a_{k}=\frac{1}{n}\sum_{k=1}^{n}\arcsin(k/n).$$ So what is the limit you are looking for?