Question

Prove 2^n > n for all n equal to or greater than 1. I mostly need help with how to solve the problem when it is greater than rather than equal to. Thanks for the help

Answer 1

If
`n` is an integer, you can use induction. First show the inequality holds for
`n=1`. You have
`2^1=2>1`, which is true.

Now assume this holds in general for
`n=k`, i.e. that
`2^k>k`. We want to prove the statement then must hold for
`n=k+1`.

Because
`2^k>k`, you have


`2^(k+1)=2*2^k>2k`

and this must be greater than
`k+1` for the statement to be true, so we require


`2k>k+1`

for
`k>1`. Well this is obviously true, because solving the inequality gives
`3k>1\implies k>\frac13`. So you're done.

If you
`n` is any real number, you can use derivatives to show that
`2^n` increases monotonically and faster than
`n`.