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?

  • $\begingroup$ I feel so stupid now! How could I not realise this earlier? Thank you so much! Now this becomes easy thanks to your hint! $\endgroup$ – Evan William Chandra Jan 11 at 6:58

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