0
$\begingroup$

Given functions $f$ and $g$ with following relation

$\exists c>0 \text{ and } n_0 \in \mathbb{N}$ s.t. $\forall n>n_0$ we have $f(n) < c\cdot h(n)$.

Prove or disprove that

$\exists c'>0 \text{ and } n_1 \in \mathbb{N}$ s.t. $\forall n>n_1$ we have $f(n+1) < c'\cdot h(n+1)$.

$\endgroup$
  • $\begingroup$ What a superb question! $\endgroup$ – John Hughes Oct 7 '17 at 2:30
  • $\begingroup$ I must be misunderstanding something in this question. It appears to me that one can take $c' = c$ and $n_1 = n_0$, and it all just holds by the first condition. I also suppose that you mean $f$ and $h$ instead of $f$ and $g$. $\endgroup$ – davidlowryduda Oct 8 '17 at 22:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.