<s>[INST] 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! :)
 [/INST] 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 $x=-3 \implies \left(\left(\frac{1}{4}\right)^n+\left(\frac{1}{3}\right)^n\right)x^n=\left(-\frac{3}{4}\right)^n+(-1)^n$
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