Radius of convergence of $\sum_{k=1}^{\infty} \frac{(2k)(2k-2)\cdots 4\cdot 2}{(2k-1)(2k-3)\cdots 3\cdot 1} z^k$ Find the radius of convergence of
$$\sum_{k=1}^{\infty} \frac{(2k)(2k-2)\cdots 4\cdot 2}{(2k-1)(2k-3)\cdots 3\cdot 1} z^k$$
Is the radius of convergence just $1$? Obviously the numerator represents even numbers and the denominator represents odd numbers. I got confused while performing the ratio test.
 A: Yes you indeed guessed the right answer. The radius of convergence is $1$. But you are probably getting confused in the working because there are a lot terms involved and you are trying to calculate everything in one step.
Instead take a step by step approach. Start by giving complicated terms convenient labels. Let us start by giving the coefficient $$\frac{\big(2k\big)\big(2k-2\big)\cdots 4\cdot 2}{\big(2k-1\big)\big(2k-3\big)\cdots 3\cdot 1}$$ of your power series, the label $\boxed{a_k}$. Then to calculate the radius of convergence by the ratio test, we need to compute the quantity $r$ (could be $\infty$) given by
$$
r = \lim_{k \to \infty}\Big|\frac{a_k}{a_{k+1}}\Big|
$$ Let us take this step by step. Firstly we already know what $a_k$ looks like. So what does $\boxed{a_{k+1}}$ look like? Well replace all of the $k$ in $a_k$ with $k+1$ to get:
\begin{align*}
a_{k+1} &= \frac{\big(2(k+1)\big)\big(2(k+1)-2\big)\cdots 4\cdot 2}{\big(2(k+1)-1\big)\big(2(k+1)-3\big)\cdots 3\cdot 1} \\
&= \frac{\big(2k+2\big)\big(2k\big)\cdots 4\cdot 2}{\big(2k+1\big)\big(2k-1\big)\cdots 3\cdot 1} \\
&= \frac{\big(2k+2\big)\boxed{\big(2k\big)\big(2k - 2\big)\cdots 4\cdot 2}}{\big(2k+1\big)\boxed{\big(2k-1\big)\big(2k-3\big)\cdots 3\cdot 1}} \\
\end{align*} What do you notice about the boxed terms above? They are the numerator and denominator of $a_k$. So you can rewrite:
$$
a_{k+1} = \frac{2k + 2}{2k+1}a_k
$$ And therefore the ratio test gives
$$
r = \lim_{k \to \infty}\Big|\frac{a_k}{\frac{2k + 2}{2k+1}a_k}\Big| = \lim_{k \to \infty}\Big|\frac{2k + 1}{2k+2}\Big| = 1
$$ as you guessed.
