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
 A: You may recall the fact that $\sin' = \cos$, which implies the standard limit
$$
\lim_{x\to 0} \frac{\sin x}{x} = \lim_{x\to 0} \frac{\sin x - \sin 0}{x- 0} = \sin' 0 = \cos 0 = 1 \tag{1}
$$
from which we have, as $1/n^2 \to 0$ as $n\to\infty$,
$$
\lim_{n\to\infty} \frac{\sin \frac{2}{n^2}}{\frac{2}{n^2}} = 1 \tag{2}
$$
Therefore, we get, for $a_n \stackrel{\rm def}{=} \sin \frac{2}{n^2}$ (for all $n\geq 1$),
$$\begin{align}
\lim_{n\to\infty} \frac{a_{n+1}}{a_n} &= 
\lim_{n\to\infty} \frac{\sin \frac{2}{(n+1)^2}}{\sin \frac{2}{n^2}}
= 
\lim_{n\to\infty} \frac{\sin \frac{2}{(n+1)^2}}{\frac{2}{(n+1)^2}}\cdot\frac{\frac{2}{(n+1)^2}}{\frac{2}{n^2}}\cdot \frac{\frac{2}{n^2}}{\sin \frac{2}{n^2}}
\\&=
\lim_{n\to\infty} \frac{\sin \frac{2}{(n+1)^2}}{\frac{2}{(n+1)^2}}\cdot\lim_{n\to\infty} \frac{n^2}{(n+1)^2}\cdot \lim_{n\to\infty} \frac{\frac{2}{n^2}}{\sin \frac{2}{n^2}}
\end{align}$$
Now, all three limits in this last expression exist and are equal to $1$ (the first and last because of $(2)$). Therefore,
$$
\boxed{\lim_{n\to\infty} \frac{a_{n+1}}{a_n} =
1}
$$
A: Another approach.
$$a_n=\sin \left(\frac{2}{n^2}\right)\implies R_n=\frac{a_n}{a_{n+1}}=\sin \left(\frac{2}{n^2}\right) \csc \left(\frac{2}{(n+1)^2}\right)$$ Now, using Taylor series
$$R_n=\frac{a_n}{a_{n+1}}=1+\frac{2}{n}+O\left(\frac{1}{n^2}\right)$$ and since $R=R_\infty$, then ...
A: $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.
