Limit of nth root of polynomial series

Question

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?

Answer 1

We'll use the squeeze theorem, but first let us find upper and lower bounds for $|P(n)| = |a_0 + a_1 n + \cdots + a_k n^k|$, assuming of course that $a_k \ne 0$. Note that $$ \lim_{n \to \infty} \frac{|P(n)|}{n^{k}} = \lim_{n\to\infty}\left|\frac{a_0}{n^{k}} + \frac{a_1}{n^{k-1}} + \cdots + \frac{a_{k-1}}{n} + a_k\right| = |a_k| > 0. $$ Hence $\frac{|P(n)|}{n^k}$ is bounded from above and from below by positive constants, i.e. $$ cn^{k} \le |P(n)| \le Cn^{k}. $$ For instance you can take $c = |a_k|/2$ and $C = 2|a_k|$ and the inequality will hold for all sufficiently large $n$. You can see this as follows: $$ \lim_{n \to \infty} \frac{|P(n)|}{n^{k}} = |a_k| $$ means by definition that for every $\epsilon > 0$ there exists a number $N$ such that $n \ge N$ implies $$ \left|\frac{|P(n)|}{n^{k}} - |a_k|\right| < \epsilon, $$ or equivalently, $$ |a_k| - \epsilon < \frac{|P(n)|}{n^{k}} < |a_k| + \epsilon $$

First pick $\epsilon = |a_k|$. Then there exists by definition a number $N_1$ such that $n \ge N_1$ implies that $$ \frac{|P(n)|}{n^{k}} < 2|a_k|. $$

Repeat the same thing with $\epsilon = |a_k|/2$ to find an $N_2$ such that $n \ge N_2$ implies that $$ |a_k|/2 = |a_k| - |a_k|/2 < \frac{|P(n)|}{n^{k}}. $$

Now for $n \ge \max\{N_1, N_2\}$, you get both inequalities.