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.
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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 $2n\geq n+1$.
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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 $$\frac1{2^{n+1}}<\frac1{n+1}$$
Parcly Taxel
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