<s>[INST] Does it converge? $\sum_{n=1}^{\infty} \arcsin^n\frac{1}{n}$ Does it converge? $\sum_\limits{n=1}^{\infty} \arcsin^n\frac{1}{n}$
I'm trying to solve it with the ratio test and using L'hopital, but it does not seem to go well. Can anyone help please?
 [/INST] It converges drastically.
Since,
for $0 \le x \le \pi/2$
we have
$1 \ge \dfrac{\sin x}{x}
\ge \dfrac{2}{\pi}
$,
so that
$1 \ge \dfrac{ x}{\arcsin x}
\ge \dfrac{2}{\pi}
$.
Therefore
$1 
\le \dfrac{\arcsin x}{ x}
\le \dfrac{\pi}{2}
$,
so
$\arcsin x
\le \dfrac{\pi x}{2}
$.
In particular,
$\arcsin \dfrac1{n}
\le \dfrac{\pi }{2n}
$.
Therefore
$\sum_\limits{n=1}^{\infty} \arcsin^{1+c}\dfrac{1}{n}
$
converges for any
$c > 1$
by comparison with
$\sum_\limits{n=1}^{\infty} \dfrac{1}{n^{1+c}}
$.
Note that
$\sum_\limits{n=1}^{\infty} \arcsin\dfrac{1}{n}
$
diverges since
$\arcsin\dfrac{1}{n}
\ge \dfrac1{n}
$.
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