Determine the interval of convergence of the series:

Question

$\sum_{k=1}^\infty \sin(\frac{\pi}{k})(x-1)^k$

I used the ratio test to reach:

$\vert(x-1)\vert\lim_{n\to \infty} \vert \frac {\sin(\frac{\pi}{k+1})}{\sin(\frac{\pi}{k})}\vert\lt1$

but I don't know where to go from here

Answer 1

HINT

Note that

$$\lim_{k\to \infty} \frac {\sin(\frac{\pi}{k+1})}{\sin(\frac{\pi}{k})}=\lim_{k\to \infty} \frac{k}{k+1}\frac {\sin(\frac{\pi}{k+1})}{\frac{\pi}{k+1}}\frac {\frac{\pi}{k}}{\sin(\frac{\pi}{k})}\to 1$$

then

$$\vert x-1\vert\ \lim_{k\to \infty} \frac {\sin(\frac{\pi}{k+1})}{\sin(\frac{\pi}{k})} \to \vert x-1\vert$$

Answer 2

We take the limit as $n$, not $x$, goes to $\infty$. This is because we want to find the radius of convergence of $x$, and if $x\to\infty$, then $R\to\infty$ which is not what we want. The limit of $a_n$ is indeterminant, so we use L'Hôpital's rule with respect to $n$

$$\displaystyle \lim_{n\to\infty}\frac{\sin{\frac{\pi}{n+1}}}{\sin{\frac{\pi}{n}}}=\lim_{n\to\infty}\left(\frac{\cos{\frac{\pi}{n+1}}}{\cos{\frac{\pi}{n}}}\frac{\frac{-\pi}{(n+1)^2}}{\frac{-\pi}{n^2}}\right)=\lim_{n\to\infty}\left(\frac{\cos{\frac{\pi}{n+1}}}{\cos{\frac{\pi}{n}}}\right)\lim_{n\to\infty}\left(\frac{\frac{-\pi}{(n+1)^2}}{\frac{-\pi}{n^2}}\right)$$

The limit of the cosines is $1$ and

$$\displaystyle\frac{\frac{-\pi}{(n+1)^2}}{\frac{-\pi}{n^2}}=\frac{1}{(1+\frac{1}{n})^2}$$

Answer 3

From what you have gotten, the sum converges for $|x-1| < 1$ or $0 < x < 2$.

The only uncertainty is at the endpoints. The key to these is noticing that $\sin(\frac{\pi}{k}) \to \frac{\pi}{k} $ as $k \to \infty$. In particular, if $0 \le x \le \pi/2$, $\frac{2x}{\pi} \le \sin(x) \le x$.

At $x=2$, the sum is $\sum_{k=1}^\infty \sin(\frac{\pi}{k}) $ and this diverges because $\sum_{k=1}^n \sin(\frac{\pi}{k}) \ge \sum_{k=1}^n \frac{\pi}{k}\frac{2}{\pi} = 2\sum_{k=1}^n \frac{1}{k} $ and this is the well-known diverging harmonic series.

At $x=0$, the sum is $\sum_{k=1}^\infty (-1)^k\sin(\frac{\pi}{k}) $ and this converges because it is an alternating series of decreasing terms.

(added after looking at Matt Werner's answer)

I realize that we have to prove that $ r_k =\frac {\sin(\frac{\pi}{k+1})}{\sin(\frac{\pi}{k})} \to 1$.

First of all, $r_k < 1$ since sine is monotonic decreasing.

There are a number of ways to show that $r_k \to 1$. One way is using $\sin'(0) = 1$.

I will try another.

By the mean value theorem, $\sin(x+h)-\sin(x) =h\cos(x+c) $ where $0 \le c \le h$. Therefore $\sin(\frac{\pi}{k})-\sin(\frac{\pi}{k+1}) =\frac{\pi}{k(k+1)}\cos(\frac{\pi}{k}+c) \le\frac{\pi}{k(k+1)} $.

Therefore, since $\sin(x) \ge \frac{2x}{\pi}$,

$\begin{array}\\ 1-\frac{\sin(\frac{\pi}{k+1})}{\sin(\frac{\pi}{k})} &\le\frac{\pi}{k(k+1)\sin(\frac{\pi}{k})}\\ &\le\frac{\pi}{k(k+1)\frac{2}{k}}\\ &\le\frac{\pi}{2(k+1)}\\ \text{so}\\ \frac{\sin(\frac{\pi}{k+1})}{\sin(\frac{\pi}{k})} &\ge 1-\frac{\pi}{2(k+1)}\\ \end{array} $