Prove that $\sum_{1}^{\infty} a_{n} \lt \infty \rightarrow \sum_{1}^{\infty} \arcsin(a_{n}) \lt \infty$ Let be $\{a_{n}\}$ a sequence of real number / $0\lt a_{n} \le 1 \; \forall n \in \Bbb N $. Prove that $\sum_{1}^{\infty} a_{n} \lt \infty \rightarrow \sum_{1}^{\infty} \arcsin(a_{n}) \lt \infty$
So, what I did:
$$\text{If} \; \sum_{1}^{\infty} a_{n} \lt \infty \; \rightarrow \lim_{x\to \infty} a_{n} = 0$$
So, by the Limit comparison test, which states that if $\lim_{x\to \infty} \frac{a_{n}}{b_{n}} \gt 0 \; \text{then} \sum_{1}^{\infty} a_{n} \lt \infty \iff \sum_{1}^{\infty} b_{n} \lt \infty$
Then, $\lim_{x\to \infty} \frac{a_{n}}{b_{n}} = \lim_{t\to 0} \frac{\arcsin(t)}{t} = \frac{d}{dt} \arcsin(t)|_{t=0}=1 \gt 0$.
So that would be it. I don´t know if it's done correctly. I don't know why the hypothesis that  $0\lt a_{n} \le 1$. This is actually a two-part exercise but I'll put the second part in another Question. Thanks in advance.
 A: Defining the arcsine function as $\arcsin(x)=\int_0^x\frac1{\sqrt{1-t^2}}\,dt$, we have the following estimate for $a_n\in (0,1)$.
$$\begin{align}
|\arcsin(a_n)|&=\int_0^{a_n}\frac1{\sqrt{1-x^2}}\,dx\\\\
&\le \frac{a_n}{\sqrt{1-a_n^2}}
\end{align}$$
Inasmuch as $a_n\to 0$ as $n\to \infty$, there exists a number $N$ such that for all $n>N$,  $\sqrt{1-a_n^2}>\frac12$.
Hence we assert that
$$\left|\sum_{n=N+1}^\infty \arcsin(a_n)\right|\le 2\sum_{n=N+1}^\infty a_n<\infty$$
and the series of interest converges.
A: Hint: $\frac {\arcsin x-x} {x^{2}} \to 0$ as $x \to 0$ by L'Hospital's Rule (applied twice). Now use the fact that $\sum a_n <\infty$ implies $\sum a_n^{2}$ is also convergent.
Note that $\arcsin x$ is not defined for $x>1$. That is the reason for assuming that $a_n \leq 1$.
A: You have a convergence criterion from  Asymptotic  Analysis for series with positive terms:

If $\:a_n,b_n >0\:$ and  $a_n\sim_{n\to \infty} b_n$, then $\sum\limits_{n=0}^\infty a_n$ converges (resp. diverges) if and only if  $\sum\limits _{n=0}^\infty b_n$ does.

Now here, as you observed, $\lim_{n\to\infty}a_n=0$, so $\:\arcsin a_n\sim_{n\to\infty}a_n$, and you can apply the above result.
