From the section on induction in Miklos Bona's combinatorics book: Prove [using induction] there exists an $N\in\mathbb{N}$ such that $n!<(\frac{n}{2.5})^n$ for all $n>N$. For what it's worth, I took this to be $$\exists N\in\mathbb{N}\forall n\in \mathbb{N}[n>N\implies n!<(\frac{n}{2.5})^n]$$ in quantifiers. I may be wrong on that and maybe that's why I had an issue with the problem (it's been a while since truth tables and all the rest of it).
It's one of the first - if not the first - 'existence by induction' proofs I've ever seen so begrudgingly I gave up and used his solutions manual (linked).
My proof closely aligned with his besides that I couldn't see how yellow implies blue, so I said: $\frac{b_{n+1}}{b_n}<\frac{2.5}{2.6}<1$ when $n>m$ implies $b_{n+1}<b_n$ when $n>m$; i.e., $b_n$ is decreasing when $n>m$. Since $b_n$ is decreasing and each $b_n>0$, the sequence converges to some $\ell$. Hence $$\ell=\lim_{n\to\infty}b_{n+1}=\lim_{n\to\infty}2.5b_nc_n=\ell\cdot\frac{2.5}{e},$$ which implies $0=\ell\cdot\frac{2.5}{e}-\ell=\ell(\frac{2.5}{e}-1).$ It follows that $\ell=0.$ So, since $b_n\to0$, there must be some $N\in\mathbb{N}$ such that when $n>N$ we have $b_n<1$; i.e., $n!<(\frac{n}{2.5})^n.$ QED.
My questions: 1. Isn't it true that both his solution and mine complete the proof, just using basic properties of sequences? Why - or where - is induction even needed? And if he somehow was using induction, where's his base case? 2. How exactly does yellow imply blue? 3. Any further reading on more problems like this (existence using induction)?