# R and $\rho$ when calculating the interval of convergence of a power series

I've got the following problem, and I am confused on what to claim the radius of convergence is:

Find the radius of convergence and the interval of convergence of each power series. (Be sure to include a check for convergence at the endpoints of the interval.) $$\sum_{n=1}^{\infty}\frac{(-1)^{n}}{\sqrt{n+1}}(4x)^{n}$$

I've calculated, using the Ratio Test, that the series will converge when $|4x|<1$, or when $|x|<\frac{1}{4}$...

\begin{align*} \lim_{n\to\infty}|\frac{a_{n+1}}{a_{n}}|&=\lim_{n\to\infty}|\frac{(4x)\cdot(4x)^{n}}{\sqrt{n+2}}\cdot\frac{\sqrt{n+1}}{(4x)^{n}}|\\ &=|4x|\cdot\lim_{n\to\infty}\sqrt{\frac{n+1}{n+2}}\\ &=|4x|\cdot 1 \end{align*} Here, $\rho=1$, $R=\frac{1}{\rho}=1$. This power series will converge for all $|4x|<1$, or $|x|<\frac{1}{4}$.

I was told in my class notes that the radius of convergence is $\frac{1}{\rho}$, which in this case is $1$... but it would seem to me that it should be $\frac{1}{4}$. Could somebody please clarify what the radius of convergence is in this context, then?

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Why don't you look up the theorem in your book. I think you're misapplying it. Perhaps $\rho$ is the limit when you leave out $x$ from your calculation, that is the limit of the coefficient. In your case, the limit of the ratio of your coefficients is 4, which is $\rho$, so that $R = 1/4$.
It is common in calculus books to work with coefficients only, no $x$. Then students get to memorize another formula, and lose even more contact with the ground. –  André Nicolas Jul 8 '13 at 2:10