(Simple question) Radius of covergence of power series? For the power series:
$$
\sum\limits_{n=1}^{\infty}\frac{(x-1)^n}{2^n}
$$
Would radius of convergence be $$ x = 1 $$ ?
 A: The center of your interval of convergence is $x=1$, but not the radius.  To find the radius of convergence you need to use the ratio test.
$$\lim_{x\to \infty}\left| \frac{(x-1)^{n+1}}{2^{n+1}}\cdot \frac{2^{n}}{(x-1)^{n}}\right|=\lim_{x \to \infty} \left|\frac{x-1}{2}\right| $$
To converge, this needs to be less than $1$.  Thus $|x-1|<2$.  So your radius of convergence is $2$ and should be centered at $x=1$.
This means your interval should contain the interval $(-1, 3)$.  You need to check the endpoints to see whether they are included or not since the ratio test in inconclusive if that limit above is actually $1$ exactly.
A: To complement the answer by CPM: You can also use the formula of Chauchy-Hadamard. It states, that the series $\sum a_n (x-c)^n$ has the radius of convergence $$R=\frac{1}{\lim \sup_{n\to\infty} \sqrt[n]{|a_n|}}$$ Thus $$R=\frac{1}{\lim \sup_{n\to\infty} \sqrt[n]{\frac{1}{2^n}}}=\frac{1}{\lim \sup_{n\to\infty} \frac{1}{2}} = 2$$
A: An often easier way is that the radius of covergence is the least upper bound of the $r\ge 0$ such that $a_nr^n\to 0$ as $n\to\infty$.
Here, $\,\dfrac{a_n}{r^n}=\Bigl(\dfrac r2\Bigr)^n$ and it is well known it tends to $0$ if and only if $\,\dfrac r2<1\iff r<2$, hence $R=2$.
