Calculating the convergence radius of a power series I've tried to calculate the convergence radius of the following power series:
$$\sum_{n=1}^{\infty}\frac{3^n+4^n}{5^n+6^n}x^n$$
The Cauchy–Hadamard theorem doesn't help in this situation (I think).
So what I did is I tried to apply the d'Alembert ratio test to it and got the following limit:
$$\lim_{n\to\infty}\frac{\frac{3^n+4^n}{5^n+6^n}}{\frac{3^{n+1}+4^{n+1}}{5^{n+1}+6^{n+1}}}=\lim_{n\to\infty}\frac{(3^n+4^n)(5^{n+1}+6^{n+1})}{(5^n+6^n)(3^{n+1}+4^{n+1})}$$
but I haven't mannaged to solve in any way. I tried to calculate the limit of the function $$\lim_{x\to\infty}\frac{(3^x+4^x)(5^{x+1}+6^{x+1})}{(5^x+6^x)(3^{x+1}+4^{x+1})}$$ but of course that Lhospital's rule doesn't help (because it's in the power of n) so I was wondering:


*

*Is there a different way to find the convergence radius by using something other than the ration test?

*Might there be a identity regarding $$a^n+b^n=?$$ or $$\frac{a^n+b^n}{a^{n+1}+b^{n+1}}=?$$


$$\begin{align}
\frac{(3^n+4^n)(5^{n+1}+6^{n+1})}{(5^n+6^n)(3^{n+1}+4^{n+1})} &=\frac{6\cdot24^n+6\cdot18^n+5\cdot20^n+5\cdot15^n}{4\cdot24^n+4\cdot20^n+3\cdot18^n+3\cdot15^n}\\
&=\frac{6+6(3/4)^n+5(5/6)^n+5(5/8)^n}{4+4(5/6)^n+3(3/4)^n+3(5/8)^n}\to\frac32
\end{align}$$
