1
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.