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 $$\frac{a^n+b^n}{a^{n+1}+b^{n+1}}=?$$


I'd go about like this:
$$\lim_{n\to\infty}\frac{\frac{3^n+4^n}{5^n+6^n}}{\frac{3^{n+1}+4^{n+1}}{5^{n+1}+6^{n+1}}}=\lim_{n\to\infty}\frac{(3^n+4^n)(5^{n+1}+6^{n+1})}{(5^n+6^n)(3^{n+1}+4^{n+1})}$$
$$=\lim_{n\to\infty}\frac{((\frac{3}{4})^n+1)(5\cdot (\frac{5}{6})^n+6)}{((\frac{5}{6})^n+1)((\frac{3}{4})^n\cdot 3+4)}$$
$$=\frac{(0+1)(5\cdot 0 + 6)}{(0 + 1)(3 \cdot 0 + 4)}$$
$$=\frac32 > 1$$
The conclusion is left for the interested reader.
