<s>[INST] 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
 [/INST] 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,
$$
\boxed{\lim_{n\to\infty} \frac{a_{n+1}}{a_n} =
1}
$$
</s>