[Math] Prove $\sum_{n=1}^\infty \arcsin({\frac 1 {\sqrt{n}}})$ is divergent.

Question

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!

Answer 1

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.