I was studying complex analysis and wanted to find the radius of convergence of the power series $$\sum_{n=1}^\infty\frac{2^n+3^n}{4^n+5^n}z^n$$ I used 'root test' and had to find the limit of the form $$\left(1+\left(\frac23\right)^n\right)^{1/n}$$ (say, it's $a_n$) It's pretty similar to the definition of $e$. And since $(\frac23)^n$ converges to $0$ more rapidly than $\frac1n$ does, I think $a_n$ must converges to $1$. Moreover, the expression like $$\lim_{n\to\infty}a_n=\lim_{n\to\infty}\left[\left(1+\left(\frac23\right)^n\right)^{(\frac32)^n}\right]^{(\frac23)^n\times\frac1n}$$ is of the form $$e^0$$ and it equals to $1$. So I can conclude that the radius of convergence is $\frac53$.
But I can't give the precise reason for $a_n$ being approaching $1$. Can anybody give me the right procedure?
A simple way to show it converges to $1$ is to note that for $a \gt 1, n \gt 1, a^{1/n} \lt a$ Then you can say $1 \lt \left(1+\left(\frac23\right)^n\right)^{1/n} \lt\left(1+\left(\frac23\right)^n\right) \to 1$