Proof by induction: $ 2^n \ge n^2$ for $n\ge4$

Question

The first part is clear but in the second I did this:

$2^{n+1}=2^n\cdot 2 \ge n^2\cdot 2=n^2+n^2=n^2+n\cdot n\ge n^2+n\left(2+\frac{1}{n}\right)=(n+1)^2$

I'm not sure if I the assumption: $n\ge 2+(1/n)$ is correct or I should prove it. Can you help me? Thanks :)

Ehm, is $2^3\ge3^2$? However the assertion is true for $n>3$; in this case $n>2+\frac{1}{n}$, because $1/n<1$. — egreg, Oct 20 '15 at 23:30

score 2 · Accepted Answer · answered Oct 20 '15 at 23:36

2

You just have to note that, for $n>3$, $$ \frac{1}{n}<1 $$ so $$ n>3>2+\frac{1}{n} $$ Yes, this should be noted in the proof.

answered Oct 20 '15 at 23:36

egreg

1 Answers1