convergence radius of $\displaystyle{\sum_{n=0}^{\infty}}(\sqrt{ 4^n +3^n}＋ (-1)^n\sqrt{ 4^n-3^n})x^n$

Question

I want to know the convergence radius of $\displaystyle{\sum_{n=0}^{\infty}}(\sqrt{ 4^n +3^n}＋(-1)^n\sqrt{ 4^n-3^n})x^n$.

Firstly, I tired to calculate $\lim_{k\to\infty}\left|\frac{a_{k}}{a_{k+1}}\right|$,but I noticed this series does not converse.So we should try to use another test to check the convergence.

Thank you in advance, my teachers.

Answer 1

One way is to use the root test, which says the radius of convergence is always $\frac{1}{\limsup_{n \to \infty} |a_n|^{1/n}}$ where $a_n$ is the coefficient of $x^n$. This works even if this root doesn't actually converge to one number.

Now it's reasonable to expect that $|a_{2n}|^{1/(2n)}$ and $|a_{2n+1}|^{1/(2n+1)}$ each converge here. So you're basically looking at $\lim_{n \to \infty} \left ( \sqrt{4^n+3^n}+\sqrt{4^n-3^n} \right )^{1/n}$ (after changing variables to $2n$) and $\lim_{n \to \infty} \left ( \sqrt{4^n+3^n}-\sqrt{4^n-3^n} \right )^{1/n}$ (after changing variables to $2n+1$). The former limit is the bigger one, which is the one that matters for the limsup, and it is $2$. So the radius of convergence is $1/2$.

Answer 2

$${\sum_{n=0}^{\infty}} \left(\sqrt{ \; \; 1 \; \; + \left( \frac{3}{4} \right)^n \; \; \; } \; \;＋ \;(-1)^n \;\sqrt{ \; \; 1 \; \; - \left( \frac{3}{4} \right)^n \; \; \; } \; \;\right)(2x)^n$$.

Answer 3

Note that your principal terms are positive.

$$ a_n = 4^{n/2} x^n(1+\dfrac{1}{2} \dfrac{3^n}{4^n}-\dfrac{1}{8}\dfrac{9^n}{16^n}+(-1)^n(1-\dfrac{1}{2} \dfrac{3^n}{4^n}+\dfrac{1}{8}\dfrac{9^n}{16^n}) + o(\dfrac{x^n9^n}{16^n})) $$

$$a_{2n}= 2 \times4^{n}x^{2n}+o(2 \times 4^{2n}x^{2n})$$ $$a_{2n+1}=x^{2n+1}3^{2n+1} +o(x^{2n+1}3^{2n+1} ) $$

Both series have to converge :

Hence

$$ R=\min(\dfrac{1}{3},\dfrac{1}{2}) =\dfrac{1}{3}$$