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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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    $\begingroup$ Beware the abuse of notation there. $\endgroup$ – Raphael May 19 '12 at 18:41
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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 $f(n)\in\Theta(f(n+1))$.
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  • $\begingroup$ oh OK then. I was wrong I thought they are Θ of each other. Thank you very much!!! $\endgroup$ – Talom May 19 '12 at 15:45
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    $\begingroup$ This (abuse of) notation suggests that numbers can be in Landau relations. I think you should properly define the two sequences behind the scenes as to not confuse anybody. $\endgroup$ – Raphael May 19 '12 at 18:41

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