(This answer is just a completion to Didier Piau'a answer). Using the comparison criterion with limit also works here.
Comparison Criterion with Limit Suppose $a_n,b_n>0$ and $\lim_{n \to \infty} \frac{a_n}{b_n}=L \in (0,\infty)$. Then $\sum a_n$ and $\sum b_n$ have the same nature. (i.e. if one converges, so is the other one; if one diverges, so is the other one)
The proof of this criterion is quite simple, and the idea is that for $\varepsilon$ small enough and $n$ large we can write
$ b_n(L-\varepsilon) and $ a_n(\frac{1}{L}-\varepsilon)
In your case, the fact that $ \displaystyle \lim_{n \to \infty} \frac{\arcsin \frac{1}{n}}{\frac{1}{n}}=1 \in (0,\infty)$ simply implies via the criterion above that $ \sum \arcsin \frac{1}{n}$ has the same nature as $\sum \frac{1}{n}$ which is well known to be divergent.