# How to formally prove that $f(n)=\Theta f(n+1)$

How to formally prove that $f(n)=\Theta f(n+1)$?

It's supposed to be easy, but I still can't get it. Thank you very much.

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Beware the abuse of notation there. –  Raphael May 19 '12 at 18:41

This depends on the sequence $(f(n))$. For example:
• If $f(n)=n!$ then $f(n)\ll f(n+1)$ hence $f(n)\notin\Theta(f(n+1))$.
• If $f(n)=\frac1{n!}$ then $f(n)\gg f(n+1)$ hence $f(n)\notin\Theta(f(n+1))$.
• If $f(n)=a^n$ with $a\ne0$ then $f(n)=\frac1af(n+1)$ hence $f(n)\in\Theta(f(n+1))$.