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.

As noted by others, there are infinite many summands, one cannot simply distribute the limit operator to them.
The following might be over-killed, but I think it is somehow interesting:
We know that 
\begin{align*}
\lim_{x\rightarrow 0}\dfrac{\sin^{-1}x}{x}=1,
\end{align*}
given $\epsilon\in(0,1)$, there is an $N$ such that 
\begin{align*}
1-\epsilon<\dfrac{\sin^{-1}x}{x}<1+\epsilon
\end{align*}
for all $n\geq N$ and $0<x<1/n$.
Note that 
\begin{align*}
\sum_{k=1}^{n}\sin^{-1}\left(\dfrac{k}{n^{2}}\right)&=\sum_{k=1}^{n}\dfrac{\sin^{-1}\left(\dfrac{k}{n^{2}}\right)}{\dfrac{k}{n^{2}}}\cdot\dfrac{k}{n^{2}}\\
&=\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{\sin^{-1}\left(\dfrac{k}{n^{2}}\right)}{\dfrac{k}{n^{2}}}\cdot\dfrac{k}{n},
\end{align*}
pluggint to the $\epsilon$-inequality for large $n$, we have
\begin{align*}
(1-\epsilon)\cdot\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{k}{n}<\sum_{k=1}^{n}\sin^{-1}\left(\dfrac{k}{n^{2}}\right)<(1+\epsilon)\cdot\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{k}{n}.
\end{align*}
Taking $n\rightarrow\infty$, the sum $\dfrac{1}{n}\displaystyle\sum_{k=1}^{n}\dfrac{k}{n}$ is simply the Riemann sum of $\displaystyle\int_{0}^{1}xdx=\dfrac{1}{2}$.
The arbitrariness of $\epsilon\in(0,1)$ gives the limit as $\dfrac{1}{2}$.
