Q:

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?

A:

Simply observe by algebraic rules for limits:
$$\left(\left|\left(\frac{3^n+\left(-4\right)^n}{5}\right)^{\frac{1}{n}}\right|\right)=\\=|-4|\cdot \left(\left|\left(\frac{{\frac{-3}{4}}^n+\left(1\right)^n}{5}\right)^{\frac{1}{n}}\right|\right) \to \\ \to 4\cdot 1=4$$
Indeed note that:
$$\left(\frac{{(-\frac{3}{4})}^n+\left(1\right)^n}{5}\right)^{\frac{1}{n}}=\frac{\left({(-\frac{3}{4})}^n+1\right)^\frac{1}{n}}{5^\frac{1}{n}}{} \to \frac11 = 1$$

Indeed:
if $\forall a_n \to 1$
$$(a_n)^{\frac1n}=e^{\frac{\log a_n}{n}}\to e^0=1$$
and
$$(5)^{\frac1n}=e^{\frac{\log 5}{n}}\to e^0=1$$

and more in general

if $\forall a_n \sim n^k$
$$(a_n)^{\frac1n}=e^{\frac{\log a_n}{n}}\sim e^{k\frac{\log n}{n}}\to e^0=1$$

EG
http://www.wolframalpha.com/input/?i=lim+(n%5E1000000000)%5E(1%2Fn)+n-%3E%2Binf

