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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! :)

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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, $R=3$.

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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

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

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  • $\begingroup$ This is the ratio test, not the root test (and yes, it works smoothly in the present case). $\endgroup$ – Did Dec 4 '18 at 22:16
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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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  • $\begingroup$ Why the downvote, something wrong? $\endgroup$ – user Dec 4 '18 at 22:21

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