Proving $\frac1{2^n}<\frac1n$ inductively

Question

I know this is probably a trivial proof but I'm struggling with it for some reason. The base case is trivial. I'm stuck at the inductive step where $$\frac1{2^{n+1}}<\frac1{n+1}$$ I'm completely stumped and I have no idea how to proceed.

Answer 1

Starting from $$\frac1{2^n}<\frac1n$$ we divide both sides by 2: $$\frac1{2^{n+1}}<\frac1{2n}$$ Since $2n>n+1$ if $n>1$: $$\frac1{2n}<\frac1{n+1}$$ Hence $$\frac1{2^{n+1}}<\frac1{n+1}$$

Answer 2

You might aswell prove $2^n>n$, right? The inductive step is

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

Note that because $n\geq1$ then indeed $2n\geq n+1$.