Proof verification for $P(n) = \Omega(n^k)$ where $k \le d$

Question

I'm currently self-reading CLRS and came across this question

Let $$P(n) = a_0n^0 + \dots + a_dn^d$$ where $a_d > 0$.

Prove that for $k \le d$, $P(n) = \Omega(n^k)$

My proof is as follows:

Consider

$$\frac{P(n)}{n^k} = \frac{a_0}{n^k} + \dots + \frac{a_{k - 1}}{n^1} + a_k + a_{k + 1}n^1 + \dots + a_dn^{k - d}$$

Since $-|x| \le x$ for all real $x$, we have

$$\frac{P(n)}{n^k}\ge -\frac{|a_0|}{n^k} -\dots - \frac{|a_{n - 1}|}{n^1} + a_k + a_{k + 1}n^1 + \dots + a_dn^{k - d}$$

Since $\frac{1}{m} \le 1$ for all $m \ge 1$, we have

$$\frac{P(n)}{n^k} \ge -(|a_0| + \dots + |a_{k - 1}|) + a_k + a_{k + 1}n^1 + \dots + a_dn^{k - d}$$

Let $A = -(|a_0| + \dots + |a_{k - 1}|) + a_k - 1$ and $Q(n) = A + a_{k + 1}n^1 + \dots + a_dn^{k - d}$. Then we have

$$\frac{P(n)}{n^k} \ge 1 + Q(n)$$

Since $Q(n)$ is a polynomial with positive leading coefficient, then it is asymptotically positive, i.e. $$\exists n_0 > 0~~~~\forall n \ge n_0 ~~~~ Q(n) > 0$$

Hence,

$$\forall n \ge n_0 ~~~~ \frac{P(n)}{n^k} \ge 1$$ $$P(n) \ge 1\cdot n^k$$

Furthermore, $1\cdot n^k \ge 0$ for all $n \ge n_0 > 0$. Hence,

$$\exists C, N > 0 ~~~~ \forall n \ge N ~~~~ 0 \le Cn^k \le P(n)$$

namely $C = 1$, $N = n_0$ and therefore

$$P(n) = \Omega(n^k)$$

by definition of $\Omega$.

The proofs I've came across for this question seem to take different approaches. As such, I'd like to check if my proof is valid and if there are any improvements I can make. Thank you!

Answer 1

The proof looks good to me, but it complicates the problem a lot. Actually,

Let $Q(n) = n^k$ for some $k \leq d$. There exists $N_0 = 1$ and $c = a_k$ such that $$ P(n) > c \cdot Q(n) $$ for any $n > N_0$ since $$P(n) - c\cdot Q(n) = a_0 + a_1n + \cdots + a_{k-1}n^{k-1} + a_{k+1}n^{k+1} + \cdots + a_dn^d$$ is positive for any $n > N_0$.