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.
 A: 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 $\dfrac{1}{2}$.
A: 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
$$\lim\limits_{n \to \infty} \sum\limits_{k=1}^n \arcsin \dfrac{k}{n^2}=\lim\limits_{n \to \infty} \sum\limits_{k=1}^n \dfrac{k}{n^2}+\lim\limits_{n \to \infty} \sum\limits_{k=1}^n O\left(\dfrac{k^3}{n^6}\right)\to \frac12+0 =\frac12$$
A: $$\lim\limits_{n \to \infty} \sum\limits_{k=1}^n \arcsin \dfrac{k}{n^2} = \lim\limits_{n \to \infty} \bigg ( \arcsin \dfrac{1}{n^2} + \arcsin \dfrac{2}{n^2} + ... + \arcsin \dfrac{n}{n^2} \bigg )=$$
$$ \lim _{n\to \infty} n\times 0 = \infty \times 0 $$
Which is undefined.
Thus you need to think something else. Try to recognize it as a Riemann's Sum and turn it into an integral.
A: to see why the answer is a half, you may combine the idea of a Riemann sum defining a definite integral with the knowledge that - near the origin - the function $\sin(x)$, and therefore also its inverse $\sin^{-1}(x)$ looks like a straight line of slope 1.
for a strictly increasing function $f$ (with suitable interval of definition) we have Young's inequality (https://en.wikipedia.org/wiki/Young%27s_inequality_for_products#Standard_version_for_increasing_functions):
$$
ab \le \int_0^a f + \int_0^b f^{-1} \le ab + |a - f^{-1}(b)||b - f(a)| \tag{1}
$$
now $ \sum\limits_{k=1}^n \arcsin \dfrac{k}{n^2}$ is a Riemann sum for the integral $\int_0^{\frac1n} \arcsin(x) dx$. substituting suitable values in (1) we have:
$$
\frac1{n^2} \le \int_0^{\frac1n} \arcsin(x) dx + \int_0^{\frac1n} \sin(x) dx \le \frac1{n^2} +|(\frac1n-b_n)(\frac1n- a_n)|
$$
where $b_n = \arcsin( \frac1n)$ and $a_n = \sin( \frac1n)$. 
so you need to show that (a) the two integrals approach each other, i.e.:
$$
\lim\limits_{n \to \infty} n^2 \bigg(\int_0^{\frac1n} \arcsin(x) dx - \int_0^{\frac1n} \arcsin(x) dx \bigg) = 0 
$$
and (b) that
$$
\lim\limits_{n \to \infty}\bigg(1 - n\arcsin({n^{-1})}\bigg)\bigg(1 - n \sin({n^{-1})}\bigg) = 0 
$$
both these conclusions are straightforward consequences of the Maclaurin expansions of $\sin$ and $\arcsin$.
