# Radius of convergence of $\sum\limits_{n=1}^\infty((\frac{1}{4})^n+(\frac{1}{3})^n)x^n$

according to wolfram alpha the radius is 3. I'm struggeling with the proof and would be very glad if someone could take a short look. Here my approach using the root test:

$$a_n:=\sqrt[n]{(\frac{1}{4})^n+(\frac{1}{3})^n}<\sqrt[n]{(\frac{1}{3})^n+(\frac{1}{3})^n}=\sqrt[n]{2(\frac{1}{3})^n}=\sqrt[n]2(\frac{1}{3})=:b_n$$

$$\implies \limsup\limits_{n\rightarrow\infty}a_n<\limsup\limits_{n\rightarrow\infty}b_n=\frac{1}{3}$$

$$\left(\implies \forall x \in \mathbb{R}: |x|<3 \implies \exists \alpha \in \mathbb{R}: \lim\limits_{n\rightarrow\infty}\sum_{n=1}^\infty b_nx^n=\alpha \right)$$

But that does not necessarily mean that $$\lim\limits_{n\rightarrow\infty}\sum_{n=1}^\infty a_nx^n$$ has the same radius of convergence, does it? The radius could still be smaller, right?

Hints to a better proof are greatly appreciated! :)

Better if you compute $$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$$. In your case, $$\frac{1/4^{n+1}+1/3^{n+1}}{1/4^n+1/3^n}\cdot \frac{3^{n+1}}{3^{n+1}} =\frac{(3/4)^{n+1}+1}{3\cdot (3/4)^n+3}\overset{n\to\infty}{\longrightarrow} \frac{1}{3}.$$

Hence, $$R=3$$.

You can use also the ratio test to get the radius of convergence, I mean, if $$a_n\neq 0$$ for all $$n\ge N$$ for enough large $$N\in\Bbb N$$ and the limit

$$\rho=\lim_{n\to\infty}\frac{|a_n|}{|a_{n+1}|}$$

exists then $$\rho$$ is the radius of convergence of the power series. In this case we have that

$$\frac{|a_n|}{|a_{n+1}|}=\frac{\frac{4^n+3^n}{4^n 3^n}}{\frac{4^{n+1}+3^{n+1}}{4^{n+1}3^{n+1}}}=12\frac{4^n+3^n}{4^{n+1}+3^{n+1}}=12\frac{1+(3/4)^n}{4+3(3/4)^n}\to\frac{12}{4}=3$$

• This is the ratio test, not the root test (and yes, it works smoothly in the present case). – Did Dec 4 '18 at 22:16

We have that

$$|x|\cdot\sqrt[n]{\left(\frac{1}{4}\right)^n+\left(\frac{1}{3}\right)^n}=|x|\cdot\frac13\sqrt[n]{\left(\frac{3}{4}\right)^n+1} \to|x|\cdot\frac13\cdot 1 = |x|\cdot\frac13<1$$

and therefore the radius of convergence is $$3$$.

Moreover note that

• for $$x=3 \implies \left(\left(\frac{1}{4}\right)^n+\left(\frac{1}{3}\right)^n\right)x^n=\left(\frac{3}{4}\right)^n+1$$

• for $$x=-3 \implies \left(\left(\frac{1}{4}\right)^n+\left(\frac{1}{3}\right)^n\right)x^n=\left(-\frac{3}{4}\right)^n+(-1)^n$$

• Why the downvote, something wrong? – user Dec 4 '18 at 22:21