This is probably a trivial question for most. I am trying to find the radius of convergence of the following series:
$$\sum_{n=0}^\infty (3^{-n} + 4^{-n})x^n$$
So the answer that was given is $\hat{R} = 3$ where $\hat{R}$ is the radius of convergence. I know that we can get this answer by simply splitting the above power series to have:
$$\sum_{n=0}^\infty 3^{-n}x^n + \sum_{n=0}^\infty 4^{-n}x^n$$
Then, calculate the radius of convergence of both series (for example using the ratio test) and take the minimum of those radii to get $\hat{R} = 3$.
My question is that, why if I do this directly, for example, using the ratio test to the original series, I don't get the radius of convergence to be $3$?
The following is my working.
Let $c_n = 3^{-n} + 4^{-n}$. The ratio test (or rather formula) tells that $\hat{R} = (\lim \sup |c_{n+1}/c_n|)^{-1}$.
$$\left|\dfrac{c_{n+1}}{c_n}\right| = \left| \dfrac{3^{-n-1} + 4^{-n-1}}{3^{-n} + 4^{-n}}\right| = \left| \dfrac{3^{n} + 4^{n}}{3^{n+1} + 4^{n+1}}\right|$$
Dividing through by $4^{n+1}$, we get:
$$\left| \dfrac{\frac{1}{4}\left(\frac{3}{4}\right)^n + \frac{1}{4}}{\left(\frac{3}{4}\right)^{n+1} + 1}\right| \longrightarrow \frac{1}{4} \quad \text{ as } \quad n \to \infty$$
So, by the ratio formula, $\hat{R} = 4$.
Anyone care to point out any mistakes?
Best Answer
It is simply because$$\frac{3^{-n-1}+4^{-n-1}}{3^{-n}+4^{-n}}\color{red}{\neq}\frac{3^n+4^n}{3^{n+1}+4^{n+1}}.$$In fact\begin{align}\lim_{n\to\infty}\frac{3^{-n-1}+4^{-n-1}}{3^{-n}+4^{-n}}&=\frac13\lim_{n\to\infty}\frac{3^{-n-1}+4^{-n-1}}{3^{-n-1}+\frac134^{-n}}\\&=\frac13\lim_{n\to\infty}\frac{1+\left(\frac43\right)^{-n-1}}{1+\frac13\left(\frac43\right)^{-n}}\\&=\frac13.\end{align}