Author says P is true for n > 2, but I got it for n ≥ 2

Question

I noticed what is the potential mistake that the author has made, but I doubt myself, even for calcutating 12x16!! Please tell whether my conclusion is correct or how is the author right about their conclusion.

problem

Mistake in Book

I think the author concluded wrong that P_n implies P_n+1 if n > 2. The author, by mistake, may have put "=" sign in the last part of solution instead of ">".

in book: (n+1)! - 3ⁿ⁺¹ = 3ⁿ(n-2)
my correction: (n+1)! - 3ⁿ⁺¹ > 3ⁿ(n-2)

My approach

Assume P_n is true for some integer n.
n! - 3ⁿ > 0
n! > 3ⁿ
n! . 3 > 3ⁿ⁺¹
as n! . a ≥ n! . 3, for all integers a ≥ 3
n! . a > 3ⁿ⁺¹, for all integers a ≥ 3
let n+1 = a
n!(n+1) > 3ⁿ⁺¹, for all integers n ≥ 2
Therefore, P_n implies P_n+1 if n ≥ 2.

I got that P_n implies P_n+1 if n ≥ 2, but author says the same for n > 2

Answer 1

$P_2$ is the statement $2! - 3^2 > 0$, which is obviously false.

You seem to be misunderstanding that the step occurring immediately after the use of the induction hypothesis, $$3^n (n+1) - 3^{n+1} = 3^n \left( (n+1) - 3^1\right) = 3^n (n-2)$$ is an algebraic identity that is true for all $n$. It is not true that $$3^n (n+1) - 3^{n+1} > 3^n (n-2).$$

What the author is doing is basically writing a chain of equalities and inequalities to establish that if $P_n$ is true, then $P_{n+1}$ is true. The result of the overall chain is that of inequality, but some intermediate steps are equalities. It's like saying, if $n > 0$, then $$10^n = 2^n \cdot 5^n > 5^n.$$ The first step is equality, and the second is inequality, so overall, $10^n > 5^n$ for $n > 0$.

Answer 2

Let $Q_n$ be the statement $P_n \implies P_{n+1}$. The author says $Q_n$ holds for $n \gt 2$ and you say that it holds for $n \geq 2$.

You are giving a stronger statement. You only need to see the case of $P_2 \implies P_3$. But that is a false implies anything statement. So you are fine with saying $Q_n$ holds for $n \geq 2$.

It is just unhelpful. In order to get to the base case for the induction, you have to go up to $n = 7$ anyway. Similarly for all the $Q_2 \cdots Q_6$.

All you need to agree on is that you get something stronger than $\forall n \geq 7, Q_n$.