Evaluating a tricky limit I'm trying to calculate the radius of convergence for a power series and I'm stuck at the following step:
$$\lim_{n\to \infty}\Bigg\lvert\left(\frac{3^n+\left(-4\right)^n}{5}\right)^{\frac{1}{n}}\Bigg\rvert$$
How would you evaluate this limit?

For $n$ odd, we see that 
\begin{align*}
\left(\dfrac{4^{2k+1}-3^{2k+1}}{5}\right)^{1/(2k+1)}&=\dfrac{3}{5^{1/(2k+1)}}\left(\left(\dfrac{4}{3}\right)^{2k+1}-1\right)^{1/(2k+1)},
\end{align*}
where
\begin{align*}
\lim_{k\rightarrow\infty}\log\left(\left(\dfrac{4}{3}\right)^{2k+1}-1\right)^{1/(2k+1)}&=\lim_{k\rightarrow\infty}\dfrac{1}{2k+1}\log\left(\left(\dfrac{4}{3}\right)^{2k+1}-1\right)\\
&=\lim_{k\rightarrow\infty}\dfrac{1}{2}\dfrac{2\cdot(4/3)^{2k+1}\log(4/3)}{(4/3)^{2k+1}-1}\\
&=\log(4/3),
\end{align*}
so 
\begin{align*}
\left(\dfrac{4^{2k+1}-3^{2k+1}}{5}\right)^{1/(2k+1)}\rightarrow\dfrac{3}{1}\cdot e^{\log(4/3)}=4.
\end{align*}
For $n$ even, the same procedure leads to the limit $4$.
