User:Ale.rossi91/sandbox


 * Proposition: $$\mathfrak{N}_R=\bigcap_{\mathfrak{p}\varsubsetneq R\text{ prime}}\mathfrak{p}.$$

Proof. Let $$r\in\mathfrak{N}_R$$ and $$\mathfrak{p}$$ be a prime ideal, then $$r^n=0$$ for some $$n\in\mathbb{Z}_{>0}$$. Thus''


 * $$r^n=r\cdot r^{n-1}=0\in\mathfrak{p},\text{ since }\mathfrak{p}\text{ is an ideal,}$$

which implies $$r\in\mathfrak{p}$$ or $$r^{n-1}\in\mathfrak{p}$$. In the second case, suppose $$r^m\in\mathfrak{p}$$ for some $$m\leq n-1$$, then $$r^{m}=r\cdot r^{m-1}\in\mathfrak{p}$$ thus $$r\in\mathfrak{p}$$ or $$r^{m-1}\in\mathfrak{p}$$ and, by induction on $$m\geq1$$, we conclude $$r^m\in\mathfrak{p},\,\forall m:0< m\leq n-1$$, in particular $$r\in\mathfrak{p}$$. Therefore $$r$$ is contained in any prime ideal and $$\mathfrak{N}_R\subseteq\bigcap_{\mathfrak{p}\varsubsetneq R\text{ prime}}\mathfrak{p}$$. Conversely, we suppose $$f\notin\mathfrak{N}_R$$ and consider the set "$\Sigma :=\lbrace J\subseteq R\mid J \text{ is an ideal and } f^m\notin J\text{ for all }m\in\mathbb{Z}_{>0}\rbrace$"which is non-empty, indeed $$(0)\in \Sigma$$. $$\Sigma$$ is partially ordered by $$\subseteq$$ and any chain $$J_1\subseteq J_2 \subseteq \dots$$ has an upper bound given by $$J=\bigcup_{i\geq1} J_i\in\Sigma$$, indeed: $$J$$ is an ideal and if $$f^m\in J$$ for some $$m$$ then $$f^m\in J_l$$ for some $$l$$, which is impossible since $$J_l\in\Sigma$$; thus any chain in $$\Sigma\ne\emptyset$$ has an upper bound and we can apply Zorn's lemma: there exists a maximal element $$\mathfrak{m}\in\Sigma$$. We need to prove that $$\mathfrak{m}$$ is a prime ideal: let $$g,h\notin\mathfrak{m},gh\in\mathfrak{m}$$, then $$\mathfrak{m}\varsubsetneq\mathfrak{m}+(g),\mathfrak{m}+(h)\notin\Sigma$$ since $$\mathfrak{m}$$ is maximal in $$\Sigma$$, which is to say, there exist $$r,s\in\mathbb{Z}_{>0}$$ such that $$f^r\in\mathfrak{m}+(g),f^s\in\mathfrak{m}+(h)$$, but then $$f^rf^s=f^{r+s}\in\mathfrak{m}+(gh)=\mathfrak{m}\in\Sigma$$, which is absurd. Therefore if $$f\notin\mathfrak{N}_R$$ then $$f$$ is not contained in any prime ideal, equivalently $$R\setminus\mathfrak{N}_R\subseteq R\setminus\bigcup_{\mathfrak{p}\varsubsetneq R\text{ prime}}\mathfrak{p}$$ and finally $$\mathfrak{N}_R\supseteq\bigcap_{\mathfrak{p}\varsubsetneq R\text{ prime}}\mathfrak{p}$$.

Proof of the theorem for the $$\cdot/\infty$$ case
Case 1: suppose $$(b_n)$$ strictly increasing and divergent to $$+\infty$$, and $$l<\infty$$. By hypothesis, we have that for all $$\epsilon/2>0$$ there exists $$\nu>0$$ such that $$\forall n>\nu$$
 * $$\left|\,\frac{a_{n+1}-a_n}{b_{n+1}-b_n}-l\,\right|<\frac{\epsilon}{2},$$

which is to say
 * $$l-\epsilon/2<\frac{a_{n+1}-a_n}{b_{n+1}-b_n}\nu.$$

Since $$(b_n)$$ is strictly increasing, $$b_{n+1}-b_n>0$$, and the following holds
 * $$(l-\epsilon/2)(b_{n+1}-b_n)\nu$$.

Next we notice that
 * $$a_n=[(a_n-a_{n-1})+\dots+(a_{\nu+2}-a_{\nu+1})]+a_{\nu+1}$$

thus, by applying the above inequality to each of the terms in the square brackets, we obtain
 * $$\begin{align}

&(l-\epsilon/2)(b_n-b_{\nu+1})+a_{\nu+1}=(l-\epsilon/2)[(b_n-b_{n-1})+\dots+(b_{\nu+2}-b_{\nu+1})]+a_{\nu+1}0$$ such that $$b_n\gneq0$$ for all $$n>n_0$$, and we can divide the two inequalities by $$b_n$$ for all $$n>\max\{\nu,n_0\}$$
 * $$(l-\epsilon/2)+\frac{a_{\nu+1}-b_{\nu+1}(l-\epsilon/2)}{b_n}<\frac{a_n}{b_n}<(l+\epsilon/2)+\frac{a_{\nu+1}-b_{\nu+1}(l+\epsilon/2)}{b_n}.$$

The two sequences (which are only defined for $$n>n_0$$ as there could be an $$N\leq n_0$$ such that $$b_N=0$$)
 * $$c^{\pm}_n:=\frac{a_{\nu+1}-b_{\nu+1}(l\pm\epsilon/2)}{b_n}$$

are infinitesimal since $$b_n\to+\infty$$ and the numerator is a constant number, hence for all $$\epsilon/2>0$$ there exist $$n_{\pm}>n_0>0$$, such that
 * $$\begin{align}

&|c^+_n|<\epsilon/2,\quad \forall n>n_+,\\ &|c^-_n|<\epsilon/2,\quad\forall n>n_-, \end{align}$$ therefore
 * $$l-\epsilon\max\lbrace \nu,n_{\pm}\rbrace=:N>0,$$

which concludes the proof. The case with $$(b_n)$$ strictly decreasing and divergent to $$-\infty$$, and $$l<\infty$$ is similar.

Case 2: we assume $$(b_n)$$ strictly increasing and divergent to $$+\infty$$, and $$l=+\infty$$. Proceeding as before, for all $$\frac32 M>0$$ there exists $$ \nu>0$$ such that for all $$ n>\nu$$
 * $$ \frac{a_{n+1}-a_n}{b_{n+1}-b_n}>\frac{3}{2}M.$$

Again, by applying the above inequality to each of the terms inside the square brackets we obtain
 * $$ a_n>\frac32 M(b_n-b_{\nu+1})+a_{\nu+1},\quad \forall n>\nu,$$

and
 * $$ \frac{a_n}{b_n}>\frac32 M+\frac{a_{\nu+1}-\frac32 Mb_{\nu+1}}{b_n},\quad \forall n>\max\{\nu,n_0\}.$$

The sequence $$(c_n)_{n>n_0}$$ defined by
 * $$    c_n=\frac{a_{\nu+1}-\frac32 Mb_{\nu+1}}{b_n}$$

is infinitesimal, thus
 * $$\forall M/2>0\,\exists \bar{n}>n_0>0\text{ such that } -M/2\bar{n},$$

combining this inequality with the previous one we conclude
 * $$  \frac{a_n}{b_n}>\frac32 M+c_n>M,\quad\forall n>\max\{\nu,\bar{n}\}=:N.$$

The proofs of the other cases with $$(b_n)$$ strictly increasing or decreasing and approaching $$+\infty$$ or $$-\infty$$ respectively and $$ l=\pm\infty$$ all proceed in this same way.

Proof of the theorem for the $$ 0 / 0 $$ case
Case 1: we first consider the case with $$l<\infty$$ and $$(b_n)$$ strictly increasing. This time, for each $$m>0$$, we can write
 * $$a_n=(a_n-a_{n-1})+\dots+(a_{m+\nu+1}-a_{m+\nu})+a_{m+\nu},$$

and
 * $$\begin{align}

&(l-\epsilon/2)(b_n-b_{\nu+m})+a_{\nu+m}=(l-\epsilon/2)[(b_n-b_{n-1})+\dots+(b_{\nu+m+1}-b_{\nu+m})]+a_{\nu+m}0$$ there are $$n^{\pm}>0$$ such that
 * $$\begin{align}

&|c^+_m|<\epsilon/2,\quad \forall m>n_+,\\ &|c^-_m|<\epsilon/2,\quad\forall m>n_-, \end{align}$$ thus, choosing $$m$$ appropriately (which is to say, taking the limit with respect to $$m$$) we obtain
 * $$l-\epsilon\max\{\nu,n_0\},$$

which concludes the proof.

Case 2: we assume $$l=+\infty$$ and $$(b_n)$$ strictly increasing. For all $$\frac32 M>0$$ there exists $$ \nu>0$$ such that for all $$ n>\nu$$
 * $$ \frac{a_{n+1}-a_n}{b_{n+1}-b_n}>\frac{3}{2}M.$$

Therefore, for each $$m>0$$
 * $$ \frac{a_n}{b_n}>\frac32 M+\frac{a_{\nu+m}-\frac32 Mb_{\nu+m}}{b_n},\quad \forall n>\max\{\nu,n_0\}.$$

The sequence
 * $$c_m:=\frac{a_{\nu+m}-\frac32 Mb_{\nu+m}}{b_n}$$

converges to $$0$$ (keeping $$n$$ fixed), hence
 * $$\forall M/2>0\,\exists \bar{n}>0\text{ such that } -M/2\bar{n},$$

and, choosing $$m$$ conveniently, we conclude the proof
 * $$\frac{a_n}{b_n}>\frac32 M+c_m>M,\quad\forall n>\max\{\nu,n_0\}.$$

Applications and examples
The theorem concerning the $$\cdot/\infty$$ case has a few notable consequences which are useful in the computation of limits.

Arithmetic mean
Let $$(x_n)$$ be a sequence of real numbers which converges to $$l$$, define
 * $$a_n:=\sum_{m=1}^nx_m=x_1+\dots+x_n,\quad b_n:=n$$

then $$(b_n)$$ is strictly increasing and diverges to $$+\infty$$. We compute
 * $$\lim_{n\to+\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\lim_n x_{n+1}=\lim_n x_n=l$$

therefore
 * $$\lim_{n\to+\infty}\frac{x_1+\dots+ x_n}{n}=\lim_{n\to+\infty}x_n.$$

Given any sequence $$(x_n)_{n\geq 1}$$ of real numbers, suppose that
 * $$\lim_{n\to+\infty}x_n$$

exists (finite or infinite), then 
 * $$\lim_{n\to+\infty}\frac{x_1+\dots+x_n}{n}=\lim_{n\to+\infty}x_n.$$

Geometric mean
Let $$(x_n)$$ be a sequence of positive real numbers converging to $$l$$ and define
 * $$a_n:=\log(x_1\cdots x_n),\quad b_n:=n,$$

again we compute
 * $$\lim_{n\to+\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\lim_n\log\Big(\frac{x_1\cdots x_{n+1}}{x_1\cdots x_n}\Big)=\lim_n\log(x_{n+1})=\lim_n\log(x_n)=\log(l),$$

where we used the fact that the logarithm is continuous. Thus
 * $$\lim_{n\to+\infty}\frac{\log(x_1\cdots x_n)}{n}=\lim_n\log\Big((x_1\cdots x_n)^{\frac{1}{n}}\Big)=\log(l),$$

since the logarithm is both continuous and injective we can conclude that
 * $$\lim_{n\to+\infty}\sqrt[n]{x_1\cdots x_n}=\lim_{n\to+\infty}x_n$$.

Given any sequence $$(x_n)_{n\geq 1}$$ of (strictly) positive real numbers, suppose that
 * $$\lim_{n\to+\infty}x_n$$

exists (finite or infinite), then 
 * $$\lim_{n\to+\infty}\sqrt[n]{x_1\cdots x_n}=\lim_{n\to+\infty}x_n.$$

Suppose we are given a sequence $$(y_n)_{n\geq1}$$ and we are asked to compute
 * $$\lim_n\sqrt[n]{y_n},$$

defining $$y_0=1$$ and $$x_n=y_n/y_{n-1}$$ we obtain
 * $$\lim_n\sqrt[n]{x_1\dots x_n}=\lim_n\sqrt[n]{\frac{y_1\dots y_{n}}{y_0\cdot y_1\dots y_{n-1}}}=\lim_n\sqrt[n]{y_n},$$

if we apply the property above
 * $$\lim_n\sqrt[n]{y_n}=\lim_nx_n=\lim_n\frac{y_n}{y_{n-1}}.$$

This last form is usually the most useful to compute limits Given any sequence $$(y_n)_{n\geq 1}$$ of (strictly) positive real numbers, suppose that
 * $$\lim_{n\to+\infty}\frac{y_{n+1}}{y_{n}}$$

exists (finite or infinite), then 
 * $$\lim_{n\to+\infty}\sqrt[n]{y_n}=\lim_{n\to+\infty}\frac{y_{n+1}}{y_{n}}.$$

Example 1

 * $$\lim_{n\to\infty}\sqrt[n]{n}=\lim_n\frac{n+1}{n}=1.$$

Example 2

 * $$\begin{align}

\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}&=\lim_n\frac{(n+1)!(n^n)}{n!(n+1)^{n+1}}\\ &=\lim_n\frac{n^n}{(n+1)^n}=\lim_n\frac{1}{(1+\frac{1}{n})^n}=\frac{1}{e}. \end{align}$$ we used the representation of $$e$$ as the limit of a sequence in the last step.

Example 3

 * $$\lim_{n\to\infty}\frac{\log(n!)}{n\log(n)}=\lim_n\frac{\log(\sqrt[n]{n!})}{\log(n)},$$

notice that
 * $$\lim_n\sqrt[n]{n!}=\lim_n\frac{(n+1)!}{n!}=\lim_n(n+1)$$

therefore
 * $$\lim_{n\to\infty}\frac{\log(n!)}{n\log(n)}=\lim_n\frac{\log(n+1)}{\log(n)}=1.$$

Example 4
Consider the sequence
 * $$a_n=(-1)^n\frac{n!}{n^n}$$

this can be written as
 * $$a_n=b_n\cdot c_n,\quad b_n:=(-1)^n,c_n:=\Big(\frac{\sqrt[n]{n!}}{n}\Big)^n,$$

the sequence $$(b_n)$$ is bounded (and oscillating), while
 * $$\lim_n\Big(\frac{\sqrt[n]{n!}}{n}\Big)^n=\lim_n(1/e)^n=0,$$

by the well-known limit, because $$1/e<1$$; therefore
 * $$\lim_{n\to\infty}(-1)^n\frac{n!}{n^n}=0.$$