I would like to prove the following series: $\sum_{n=1}^\infty {\frac{\sin{\frac{\pi}{n}}}{n}}\cdot(x-2)^n $ is convergent I would like to prove the following series: $$\sum_{n=1}^\infty {\frac{\sin{\frac{\pi}{n}}}{n}}\cdot(x-2)^n $$ is convergent (absolutely?) or divergent.
Now I know that it converges for: $|{x-2}|<1$, but I am having trouble proving it. Any ideas would be greatly appreciated.
 A: To do this we want to apply the ratio test to find the radius of convergence. So what we have is
$$\lim_{n\to \infty}\left|\frac{\frac{\sin\left(\frac{\pi}{n+1}\right)}{n+1}(x-2)^{n+1}}{\frac{\sin\left(\frac{\pi}{n}\right)}{n}(x-2)^{n}}\right|<1$$
Which simplifies immediately to
$$\lim_{n\to \infty}\left|\frac{n\sin\left(\frac{\pi}{n+1}\right)(x-2)}{(n+1)\sin\left(\frac{\pi}{n}\right)}\right|<1$$
Then we know that as $n\to \infty$, we have $\sin\left(\frac{\pi}{n}\right)=\frac{\pi}{n}$. This is because the inside of the sine function approaches zero, and for small $x$, $\sin(x)\approx x$. So then what we have is
$$\lim_{n\to \infty}\left|\frac{n\left(\frac{\pi}{n+1}\right)(x-2)}{(n+1)\frac{\pi}{n}}\right|<1$$
Which we then simplify down to
$$\left|\lim_{n\to \infty}\frac{n^2}{(n+1)^2}\right|\cdot |x-2|=|x-2|<1$$
Which leaves us with the fact that the series converges when $x \in (1,3)$, and now all thats left is to test the endpoints. For $x=1$ we have
$$\sum_{n=1}^{\infty}\frac{\sin(\frac{\pi}{n})}{n}(-1)^n$$
which converges by the alternating series test. And for $x=3$ we have
$$\sum_{n=1}^{\infty}\frac{\sin(\frac{\pi}{n})}{n}\le \sum_{n=1}^{\infty}\frac{\pi}{n^2}$$
which converges by the comparison test. So we find that the interval of convergence is $x\in [1,3]$.
A: Hint: $0 \le \sin (\pi/n)\le \pi/n.$
A: $$\left|\frac{\sin(\frac{\pi}{n})}{n}(x-2)^n\right|<\left|(x-2)^n\right|$$
The geometric series $\; \sum_{n=1}^{\infty}{(x-2)^n}\;$converges as you already found out for $|x-2|<1\;$ Hence your series converges (by applying Weierstrass M-Test).
