Let $P(n)$ be a polynomial of positive degree. Show that, $$\lim_{n\to \infty} \sqrt[n]{ |P(n)|} =1 $$
Here is my attempt, but it's clunky and I'm not 100% sure it if works or not, but here goes (apologies in advanced for formatting);
$$\text{For c > 0,} \ \lim_{n\to\infty}c^{1/n} = 1\quad\text{and}\quad\lim_{n\to\infty}n^{1/n} = 1$$
$$P(n) = a_0+a_1n+a_2n^2+\dots+a_kn^k$$ $$|a_0+a_1+a_2+\dots+a_k|\leq|P(n)|\leq |a_0+a_1n+a_2n^2+\dots+a_kn^k|\quad \text{for all}\ n\geq N$$ $$|a_0+a_1+a_2+\dots+a_k|^{1/n}\leq|P(n)|^{1/n}\leq |a_0+a_1+a_2+\dots+a_k|^{1/n}\cdot n^{k/n} $$ $$ \text{Let}\ C = a_0+a_1+a_2+\dots+a_k $$ $$|C|^{1/n}\leq |P(n)|^{1/n}\leq |C|^{1/n}\cdot n^{k/n}$$ $$1\leq |P(n)|^{1/n}\leq 1\cdot n^{k/n}$$ $$1\leq P(n) \leq 1\cdot 1^k $$
Thus, using squeeze theorem it has been demonstrated that $\lim_{n\to\infty}\sqrt[n]{|P(n)|}$. Questions: the series $|a_0+a_1+a_2+\dots+a_k|$ is assumed to converge to a constant C. Is this a fair assumption, or would the case that $|a_0+a_1+a_2+\dots+a_k|$ diverges also need to be considered? If it does, the inequality $\infty\leq P(n)\leq\infty\cdot n^{k/n}$ is invalid and squeeze theorem cannot be applied. How is it possible to go around this problem?