Greatest lower bound of the set : $\{(e^n + 2^n)^\frac1n\ | \; n\in \mathbb{N}\}$

Question

Find the greatest lower bound of the set : $\{(e^n + 2^n)^\frac1n \mid n\in \mathbb{N}\}$

I will find the limit of $a_n := (e^n + 2^n)^\frac1n $ which if exists, must equal the greatest lower bound ($\liminf$) $$ (e^n + 2^n)^\frac1n = e\left(1+\left(\frac2e\right)^n\right)^{\frac1n} \to e $$ So the required answer is $e$? is this correct?

Answer 1

Yes, that is correct. You might also want to add that $(e^n + 2^n)^{\frac{1}{n}}$ is strictly decreasing $\mathbb{N}$.