Find Radius of Convergence of $\sum_{n=0}^\infty \frac{1}{2^n}\left(x-\pi\right)^n$ This is not a homework problem (I'm on break, so time for my own studies).
Find the radius of convergence of
\begin{align}
\sum_{n=0}^\infty \frac{1}{2^n}\left(x-\pi\right)^n.
\end{align}
I have found
\begin{align}
\lim_{n\rightarrow\infty}\left|\frac{\displaystyle\frac{\left(x-\pi\right)^n}{2^n}}{\displaystyle\frac{\left(x-\pi\right)^{n+1}}{2^{n+1}}}\right|=\lim_{n\rightarrow\infty}\left|\frac{2^{n+1}\left(x-\pi\right)^n}{2^n\left(x-\pi\right)^{n+1}}\right|=\lim_{n\rightarrow\infty}\frac{2}{\left(x-\pi\right)}=\frac{2}{x-\pi}.
\end{align}
Is this right?
Thank you for your time,
 A: You haven't found the radius of convergence -- it's a constant, not variable. The radius of convergence $r$ of a power series $\sum_{n = 1}^\infty a_n(x - x_0)^n$ is such that $1/r = \lim_{n\to \infty} |a_n|^{1/n}$, assuming the limit exists. In your case, $x_0 = \pi$ and $a_n = 1/2^n$. Thus
$$\frac{1}{r} = \lim_{n\to \infty} |a_n|^{1/n} = \lim_{n\to \infty} \frac{1}{2} = \frac{1}{2}.$$
Hence, $r = 2$.

Taking another look at your work, it seems like you attempted to compute directly the radius of convergence using $r = \lim_{n\to \infty} |a_n/a_{n+1}|$, but in that case your choice of $a_n$ is incorrect -- it should be $1/2^n$, the $n$th coefficient of your power series. Then $|a_n/a_{n+1}| = 2$ for all $n$, which implies $\lim\limits_{n\to \infty} |a_n/a_{n+1}| = 2$. So again, $r = 2$.
A: You did it upside down ;). Just invert your final answer, which is correct. Solve for the radius...
$$-1 \le {{x-\pi} \over {2}} \le 1$$
$$-2 \le {x-\pi} \le 2$$
$$ \pi-2 \le x \le \pi+2$$
Looks like the radius is $2$.
A: You are trying to use the ratio test, but you have the formula inverted. It should be $\lim_{n\to \infty} \frac {|a_{n+1}|} {|a_n|} $. As in, the $|a_{n+1}|$ term should be on $\textit {top}.$
Zach's answer is not correct. The radius of convergence is not $\pi$. Rather, the power series is $\textit centered$ at $\pi$. The interval of convergence (which is the center plus/minus the radius of convergence) is $[2-\pi, 2+\pi]$ (make sure to check the endpoints for convergence as well).
