Radius of convergence of $\sum (3^{-n} + 4^{-n})x^n$

Question

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 $\hat{R} = 3$.

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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?

Answer 1

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}

Answer 2

Mistake is in the expression of $\left|\frac{c_{n+1}}{c_n}\right|$. Following is the correct one. $$\left|\frac{c_{n+1}}{c_n}\right|=\frac{3^{-n-1}+4^{-n-1}}{3^{-n}+4^{-n}}$$ Multiply numerator and denominator by $3^{n+1}\times4^{n+1}$ to get $$\left|\frac{c_{n+1}}{c_n}\right|=\frac{4^{n+1}+3^{n+1}}{3\times4^{n+1}+4\times3^{n+1}}$$ Check that now the limit evaluates correctly.

Answer 3

I would prefer to write $$\dfrac{c_{n+1}}{c_n} = \dfrac{3^{-n-1} + 4^{-n-1}}{3^{-n} + 4^{-n}} = \frac{3^{n+1}+4^{n+1}}{12 \left(3^n+4^n\right)}=\frac{1}{12} \left(4-\frac{1}{\left(1+\frac{4}{3}\right)^n}\right)$$

Answer 4

$$ \left| \dfrac{3^{-n-1} + 4^{-n-1}}{3^{-n} + 4^{-n}}\right| \neq \left| \dfrac{3^{n} + 4^{n}}{3^{n+1} + 4^{n+1}}\right| \text{.} $$

Let's try verifying it with $n = 2$.

$$ \frac{91}{300} = \left| \dfrac{3^{-3} + 4^{-3}}{3^{-2} + 4^{-2}}\right| \neq \left| \dfrac{3^{2} + 4^{2}}{3^{3} + 4^{3}}\right| = \frac{25}{91} \text{.} $$

Answer 5

You make a mistake when you writte $\frac{c_{n+1}}{c_{n}}$

$\frac{c_{n+1}}{c_{n}} = \frac{3^{-n-1} + 4^{-n-1}}{3^{-n} + 4^{-n}}= \frac{1 + (\frac{3}{4})^{n+1}}{3 + 3(\frac{3}{4})^{n}} \to \frac{1}{3}$

Answer 6

Just like you divide by $4^{n+1}$ when the powers are positive, you should multiply by $3^n$ (or $3^{n+1}$) when the powers are negative: $$\left|\dfrac{c_{n+1}}{c_n}\right| = \left| \dfrac{3^{-n-1} + 4^{-n-1}}{3^{-n} + 4^{-n}}\cdot \frac{3^n}{3^n}\right| = \left| \dfrac{3^{-1} + (1/4)(3/4)^{n}}{1 + (3/4)^{n}}\right|\to 3^{-1}=\frac13.$$