Prove $\sum_{n=1}^\infty \arcsin({\frac 1 {\sqrt{n}}})$ is divergent. 
Evaluate if the following series is convergent or divergent: $\sum\limits_{n=1}^\infty \arcsin({\frac 1 {\sqrt{n}}})$.

I think I could apply the integral test that would take me to a complex integral computation. I have checked the solution and the series are divergent.
I am trying to employ  Weierstrass comparison theorem via the maximization of the function.
Question:
How could I prove the series diverge using another test? 
Thanks in advance!
 A: Since we have $$\sin x \le x$$
We get $$\arcsin x \ge x$$ therefore, $$ \arcsin \left( \frac{1}{\sqrt n} \right) \ge \frac{1}{\sqrt n} \ge \frac 1n $$ 
Since $\displaystyle \sum_{n=1}^{\infty} \frac 1n $ diverges, given sum diverges too.
A: Note that
$$ \arcsin \left( \frac{1}{\sqrt n} \right) \sim \frac{1}{\sqrt n}$$
then the given series diverges by limit comparison test with $\sum \frac{1}{\sqrt n}$.
A: Since $\arcsin'0=1$ and $\arcsin''x>0,$ for $x>0,$ you get $\arcsin x>x$ for $x>0,$ and thus $\arcsin\dfrac 1 {\sqrt n} > \dfrac 1 {\sqrt{n}}.$ Thus
$$
\sum_{n=1}^\infty \arcsin\frac 1 {\sqrt n} \ge \sum_{n=1}^\infty \frac 1 {\sqrt n} = +\infty.
$$
What first alerts you to all this is of course that you remember what the graph of the arcsine function looks like.
A: More generally,
$\sum\limits_{n=1}^\infty \arcsin({\frac1 {n^a}})
$
converges for $a > 1$
and diverges for
$0 < a \le 1$.
This is because
$\frac{2}{\pi}x
\le \sin(x)
\le x$
for
$0 \le x \le \frac{\pi}{2}$.
Therefore
$x \le \arcsin(x)
\le \frac{\pi}{2} x
$
for
$0 \le x \le 1$.
Therefore
$\sum\limits_{n=1}^\infty \arcsin({\frac1 {n^a}})
$
converges
if and only if
$\sum\limits_{n=1}^\infty {\frac1 {n^a}}
$
converges.
A: Another way.-Since $\arcsin(x)=x+\dfrac{x^3}{6}+\dfrac{3x^5}{40}+O(x^7)$  and $0\le \dfrac {1}{\sqrt n}\le 1$ do you have $$\sum_{n=1}^\infty \arcsin\frac 1 {\sqrt n}\gt\sum_{n=1}^\infty(\dfrac {1}{\sqrt n})\gt\sum_{n=1}^\infty \dfrac 1n\to \infty $$
