Find convergence interval of $f(x) =\sum_{n=1}^∞ \left(\sin\frac{2}{n^2}\right)x^n $ $$f(x) =\sum_{n=1}^∞ \left(\sin\frac{2}{n^2}\right)x^n $$
so looks like what i have to do is use the formal 
$$\lim_{n\to∞}\left|\frac{a_{n+1}}{a_n}\right|=a$$ with $a_n=\sin\frac{2}{n^2}$. And then $R=1/a$. 
But I having trouble with the lim. Thank you for your help

$f(x) =\sum_{n=1}^∞ \left(\sin\frac{2}{n^2}\right)x^n
$
Since
$\sin(x) \approx x$
for small $x$,
$\sin\frac{2}{n^2}
\approx \frac{2}{n^2}
$
so the sum behaves like
$\sum_{n=1}^∞ \frac{2x^n}{n^2}
$
and this converges for
$-1 \lt x \lt 1$
by  comparison with
$\sum x^n$
and converges for
$x = \pm 1$
because it becomes
$\sum \frac{2}{n^2}
$
at $x=1$
and
$\sum \frac{2(-1)^n}{n^2}
$
at $x=-1$,
both of which converge.
