What's wrong in my calculation of $\lim\limits_{n \to \infty} \sum\limits_{k=1}^n \arcsin \frac{k}{n^2}$ I have the following limit to find:
$$\lim\limits_{n \to \infty} \sum\limits_{k=1}^n \arcsin \dfrac{k}{n^2}$$
This is what I did:
$$\lim\limits_{n \to \infty} \sum\limits_{k=1}^n \arcsin \dfrac{k}{n^2} = \lim\limits_{n \to \infty} \bigg ( \arcsin \dfrac{1}{n^2} + \arcsin \dfrac{2}{n^2} + ... + \arcsin \dfrac{n}{n^2} \bigg )$$
$$ \hspace{.8cm} = \arcsin 0 + \arcsin 0 + ... + \arcsin 0 $$
$$= 0 + 0 + ... + 0 \hspace{2.9cm}$$
$$=0 \hspace{5.2cm}$$
However, my textbook claims that the actual answer is in fact $\dfrac{1}{2}$. I don't see how I could reach this answer.

We can't add infinitely many terms in this way, as for example for $\sum \frac 1n$ which as we know diverges.
We have that
$$\arcsin \dfrac{k}{n^2} =\dfrac{k}{n^2}+O\left(\dfrac{k^3}{n^6}\right)$$
and therefore by Faulhaber's formula
$$\lim\limits_{n \to \infty} \sum\limits_{k=1}^n \arcsin \dfrac{k}{n^2}=\lim\limits_{n \to \infty} \sum\limits_{k=1}^n \dfrac{k}{n^2}+\lim\limits_{n \to \infty} \sum\limits_{k=1}^n O\left(\dfrac{k^3}{n^6}\right)\to \frac12+0 =\frac12$$
