Is this proof valid? The claim is $2^{k} < (k+1)!$ for $k \geq 2$

Question

Hey guys so I think I have completed this proof but I'm not sure if its valid. Here it is:

Prove that $ 2^n < (n+1)! \quad\text{for}\quad n >= 2 $

Here is my proof:

Base Case P(2) = $ 4 < 6 $

Inductive Hypothesis (IH) P(k) = $ 2^k < (k+1)! $

Proof P(k+1) = $ 2(k+1) < (k+2)! $

$ 2 * 2^k < (k+1)! * (k+2) $

I have already shown that $ 2^k < (k+1)! $ is true by IH. With $2$ being multiplied on the left and $(k+2)$ being multiplied on the right, if I can prove that $2 < (k+2) $ than the whole equation is true.

2 is always less than $k+2$ because k must be greater than or equal to 2 so the equation at minimum is $2 < 4$ .

End Proof

Is this valid? And if not, what am I missing? why is this approach wrong?

It's all true, nothing seems missing. Just the base case is P(2). — peter.petrov, May 07 '16 at 19:34
Oh woops I'll fix the base case. But I mean is the logic of proving $ 2 < (k+2) $ sound and reliable for proving these kinds of things? @peter.petrov — Laugh7, May 07 '16 at 19:39
$2<k+2$ is equivalent to $0<k$, while it's given that $2 \leq k$. So it's fine. — peter.petrov, May 07 '16 at 19:40
Typo, third line after here is my proof, you want $2^{k+1}$, not $2(k+1)$. Also, I don't like $P(k)=2^k\lt (k+1)!$. — André Nicolas, May 07 '16 at 19:52

score 1 · Answer 1 · answered May 08 '16 at 14:53

Several of the important pieces of the inductive proof are there, but everything could be written up much more smoothly. For example, your inductive hypothesis should be to "fix some $k\geq2$ and assume that $P(k) : 2^k<(k+1)!$ holds." This addresses Andre's comment--he "did not like" $P(k)=2^k<(k+1)!$ because that is technically incorrect. Why? Well, $P(k)$ is supposed to denote a proposition or a statement, not an equivalence.

That being said, I think you could more effectively write up the main part of your induction proof as follows: \begin{align} 2^{k+1}&=2\cdot2^k\tag{law of exponents}\\[1em] &< 2\cdot(k+1)!\tag{by induction hypothesis}\\[1em] &< (k+2)\cdot(k+1)!\tag{since $k\geq2$}\\[1em] &= (k+2)!\tag{by definition} \end{align}

Is this proof valid? The claim is $2^{k} < (k+1)!$ for $k \geq 2$

Here is my proof:

End Proof

1 Answers1