Prove $n! \geq 2^{n - 1}$ ; $n\geq 0$

Question

When $n=0$

$1>\frac{1}{2}$ (true)

Then I prove that $$2×n!\geq2^n$$ And $$(n+1)\times n!\geq2^n$$ So I have to prove that $n+1\geq2$

$n\geq 0$

$n+1\geq 1$ (not $2$). Could someone point out what is wrong with my reasoning please?

Using only induction for the proof is a must?or can we show it using other way — Archis Welankar, Jun 23 '19 at 15:41
To do without induction $1\times 2 \times 3 \times \dots \times n \lt 1\times 2\times 2\dots \times 2$ (same number of factors each side) and count the number of factors. But you need care with the small values of $n$. [really, the counting conceals the induction rather than making it unnecessary] — Mark Bennet, Jun 23 '19 at 15:46
is it $2\times n!$ or $(2n)!$? and have you read this: https://math.stackexchange.com/questions/76946/prove-the-inequality-n-geq-2n-by-induction?rq=1 At least you should have gone through it. — Kumar, Jun 23 '19 at 15:56

score 1 · Answer 1 · answered Jun 23 '19 at 15:34

1

Well, for $n=1$, $1!=1$ and $2^{1-1}=2^0=1$. So the strict inequality $>$ does not hold.

answered Jun 23 '19 at 15:34

Wuestenfux

21,302

It is greater or equal – Aysar Alshaar Jun 23 '19 at 15:37
Should be $\geq$. – Wuestenfux Jun 23 '19 at 15:41
I correct it now – Aysar Alshaar Jun 23 '19 at 15:47

score 0 · Answer 2 · answered Jun 23 '19 at 17:32

0

Your proof works except for the base case. You need to separately consider $n=1$, then your induction proof goes through since every successive factorial is at least double the previous one (observe that for $n\geq 1$, $n+1\geq 2$).

answered Jun 23 '19 at 17:32

Matt Samuel

59,287

DDS · Answer 3 · 2019-06-23T18:54:50.817

You should always attempt to set up Proof by Induction problems in a more formal and organized manner; for example,

$\underline{n = 0.}$

$$ 0! = 1 \geq 2^{0-1} = \frac{1}{2}. (true)$$

$\underline{n = k.}$ (This is the induction hypothesis) Assume that

$$ k! \geq 2^{k-1}. $$

$\underline{n = k + 1.}$ We must show $(k+1)! \geq 2^{k}.$

$$ (k+1)! = (k+1)k! \geq (n)2^{k} \geq (1)2^{k} = 2^{k}.$$

(as was to be shown)

Prove $n! \geq 2^{n - 1}$ ; $n\geq 0$

3 Answers3