Does the nth root of $n \cdot 2^{3n}+3^{2n}$ converge as $n \to \infty$?

Question

Consider the sequence

$a_n = \sqrt[n]{n \cdot 2^{3n}+3^{2n}}$, $n\in\mathbb{N}$.

With a string of inequalities, one can show that $a_n$ is bounded and the graph of the function $ f(x) = \sqrt[x]{x \cdot 2^{3x}+3^{2x}}$ suggests that $f$ is monotone, but how could one prove convergence and a calculate the limit of $a_n$?

I would be grateful for any help!

Answer 1

Rewrite as $\lim 8\sqrt[n]{1+(9/8)^n}\ge \lim 8\sqrt[n]{(9/8)^n}=9$ and $$\lim 8\sqrt[n]{n+(9/8)^n}\le \lim 8\sqrt[n]{2(9/8)^n}=9.$$ So the limit is $9$.

Answer 2

Note that $$a_n = \sqrt[n]{n\cdot 8^n+9^n} = 9\cdot \sqrt[n]{n\cdot \left(\frac{8}{9}\right)^n+1}$$ Denote by $ \displaystyle b_n = n\cdot \left(\frac{8}{9}\right)^n$. Since $$\frac{b_{n+1}}{b_n} = \frac{n+1}{n}\cdot \frac{8}{9} \longrightarrow \frac{8}{9}<1$$ it follows that $b_n \longrightarrow0$, and thus $a_n \longrightarrow 9$.

Addendum: Since $1\le b_n+1$ it follows that $1\le \sqrt[n]{b_n+1}\le b_n+1$. Thus, by the squeezing principle, $\sqrt[n]{b_n+1}$ tends to $1$, as I claimed.