Somemathhere

Consider sequences of natural numbers. We say that a sequence $a$ majorizes $b$, notation $a\succ b$, if $$\lim_{n\to \infty}\frac{b_n}{a_n}=0.$$ We call a set of sequences $C$ cofinal if for any sequence $v$ there is $u\in C$ such that $u$ majorizes $v$. $\textbf{Statement}.$ Every cofinal set is uncountable. $\textbf{Proof}.$ The proof is utilization of Cantor's diagonal argument. Assume $C$ is a cofinal countable set, denote its elements as $a^k$ for $k=1,2,3,\ldots$. First of all observe that $\succ$ is a transitive relation and for any finite set of sequences, there is a sequence $u$ which majorizes all of them (consider $n$ times the pointwise maximum). Construct a sequence of sequences $b^k$ as follows - take $b^1_n=\max\{a^1_1,\ldots,a^1_n\}$. Suppose we have constructed $b^j$ for $j=1,\ldots,k$. Construct $b^{k+1}$ as an nondecreasing sequence which majorizes all of $a^i$ for $i=1,\ldots,k+1$ and $b^j$ for $j=1,\ldots,k$. Thus $b^j\succ a^j$ and $b^{j+1}\succ b^j$ a

An ordinal $\alpha$ is limit ordinal if and only if there exists ordinal $\beta$ such that $\alpha=\omega\beta$. Proof. $\Leftarrow)$ If $\beta=1$ then $\alpha=\omega$ so it is limit. Assume $\omega\gamma$ is limit for all $\gamma<\beta$. We will prove that $\omega\beta$ is limit. First assume $\beta$ is limit. Then $\omega\beta=\sup\{\omega\gamma \ \big| \ \gamma<\beta\}$. Let $\theta<\omega\beta$, then $\theta<\omega\gamma'$ for some $\gamma'<\beta$. But $\omega\gamma'$ is limit thus $\theta+1<\omega\gamma'$ so $\theta+1<\omega\beta$ hence $\omega\beta$ is limit. Now let $\beta=\beta'+1$ for some $\beta'$. Thus $\omega\beta=\omega\beta'+\omega$. If $\theta<\omega\beta=\omega\beta'+\omega=\sup\{\omega\beta'+n\ \big| \ n<\omega\}$ then $\theta<\omega\beta'+n_0$ for some $n_0$. And now $\theta+1<\omega\beta'+n_0+1<\sup\{\omega\beta'+n\ \big| \ n<\omega\}$ hence $\omega\beta$ is limit. $\Rightarro

$$\displaystyle \sum_{k=1}^n(-1)^{n-k}\binom{n}{k}k^j=\begin{cases} n!, & j=n\\ 0, & j=1,2,\ldots,n\end{cases}$$ Proof. Consider the number of surjective functions $f:\{1,2,\ldots,j\}\to\{1,2,\ldots,n\}$. If we count them using the inclusion-exclusion principle we get the sum in the problem. On the other hand this number is trivially $n!$ if $j=n$ and $0$ if $j<n$. Remark 1. For $j>n$ we again find the number of surjections, which has a closed form form some $j$. Remark 2. Interesting identity which represents the above identities in a unified manner is the following $$\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}(x+k)^n=n!$$ for all $x$ (in $\mathbb{R}$ for example, but any commutative ring suffices).