Find convergence interval of $f(x) =\sum_{n=1}^∞ \left(\sin\frac{2}{n^2}\right)x^n $

Question

$$f(x) =\sum_{n=1}^∞ \left(\sin\frac{2}{n^2}\right)x^n $$

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

Answer 1

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, $$ \boxed{\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = 1} $$

Answer 2

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 ...

Answer 3

$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.