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Task is to show that

There exists some $n_{0} \in \mathbb N $, $\forall n \in \mathbb N $

$ (n_{0} \ge n \Rightarrow n! - n^4 \ge n^2 - 11n)$

Where to start with this?

By Induction over $n$, assuming $n_{0} =1:$

Holds for $n_{0} =1:$

Assume $ k! - k^4 \ge k^2 - 11k $

Show: $ (k+1)! - (k+1)^4 = ....$

Do not want the answer just headsup on the approach.. Thanks.

Anton Vrdoljak
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SIMON P
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  • The arrow in the main statement looks a bit confusing: Did you mean: There exists $n_0\in\mathbb{N}$ such that $\forall n \in \mathbb{N}$ with $n\ge n_0$ there holds $n! -n^4 \ge n^2 - 11n$? – DominikS Sep 28 '23 at 08:58
  • Prove that $n! >(n-4)^{5}$ and that $(n-4)^{5}>n^{4}+n^{2}-11n$ for $n$ sufficiently large. – geetha290krm Sep 28 '23 at 08:59
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    $n_0 = 1$ will not work, note that the inequality is not true for $n=3$. You'd want some $n_0$ larger than that; compute the terms for the first few $n\in\mathbb N$, to get a feeling for reasonable $n_0$. – DominikS Sep 28 '23 at 09:02
  • Hi, welcome to Math SE. Hint: to show $k!\ge k^4+k^2-11k\implies (k+1)!\ge(k+1)^4+(k+1)^2-11(k+1)$, you only need $k^4+k^2-11k\ge(k+1)^3+k-10$. Beyond what point do all $k$ satisfy this? – J.G. Sep 28 '23 at 09:02

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