Show that $2^n < n!$ for every positive integer $n$ with $n\geq 4$.

Question

Using Mathematical induction prove the above proposition.

Basis step can be verified easily. But how can i show that it is true for $p(n+1)$.

How do you get from $2^n$ to $2^{n+1}$? What about from $n!$ to $(n+1)!$? — Jonathan Y., Jan 29 '14 at 16:20
http://math.stackexchange.com/questions/247335/prove-formally-that-log-2-n-ge-n-for-all-integers-n3 — , Jan 29 '14 at 16:20
You could see http://math.stackexchange.com/questions/409609/how-to-prove-4n2n2-for-n-geq-4-with-induction?rq=1 and divide by 4 — Ross Millikan, Jan 29 '14 at 16:23

score 2 · Accepted Answer · answered Jan 29 '14 at 16:22

2

$$2^{n+1}=2*2^n\\\le2*n!\le(n+1)n!=(n+1)!$$

answered Jan 29 '14 at 16:22

Semsem

score 1 · Answer 2 · edited Jan 29 '14 at 16:50

Well, for induction, you usually end up proving the $n=1$ (or in this case $n=4$) case first. You've got that done.

Then you need to identify your inductive hypothesis: e.g.

$$ n!>2^n$$

and

$$ n\ge 4$$

In class the proof might look something like this:

$$(n+1)!=n!(n+1)$$

from the inductive hypothesis we have

$$ n!(n+1)>2^n(n+1)$$

since

$$n>1$$

we have

$$ 2^n(n+1)>2^n\cdot 2$$

and

$$ 2^n\cdot 2=2^{n+1}$$

Now, we can string it all togther to get the inequality:

$$ (n+1)!=n!(n+1)>2^n(n+1)>2^n\cdot 2=2^{n+1}$$

$$ (n+1)!>2^{n+1}$$

In general, it's worth trying to figure out wether it 'safe' to multiply

$$ n!>2^n$$

by

$$ n+1>2$$

while preserving the inequality.

2 Answers2