$\Theta \rho ϵ\eta \Pi \alpha \tau \pi$

Suppose for the sake of contradiction that that was not true, so that for every $N$ we had some $n\ge N$ such that $a_n=0$, let's derive a contradiction.

Since $\left(a_n\right)\to L$ then specifically for $ϵ=\frac{L}{2}$ we have some $N^{\prime }\in {\mathbb{N}}_0$ such that for every $n\ge N^{\prime }$ we have $|a_n-L|\le \frac{L}{2}$, therefore by our contradiction assumption we also know there is some $k\ge N^{\prime }$ such that $a_k=0$ so therefore because the limit exists we have $|a_k-L|\le \frac{L}{2}$ but $a_k=0$ so that $|L|\le \frac{L}{2}$ which is impossible so therefore we have a contradiction and so the original statement holds true.

Let $ϵ\in {ℝ}^{+}$ we need to pick an $N\in {\mathbb{N}}_0$ such that for every $n\ge N$ we have that $|b_n-L|<ϵ$.

Since ${\lim }_{n\to \infty }{a}_n=L$ then we can set ${ϵ}^{\prime }=f\left(ϵ\right)$ for any function $f:{ℝ}^{+}\to {ℝ}^{+}$ of our choosing, to get some $N^{\prime }$ such that for any $n\ge N^{\prime }$ we have $|a_n-L|<f\left(ϵ\right)$

Morover note that for this given $ϵ$ we can always find a $K\in {\mathbb{N}}_1$ such that $\frac{1}{K}<g\left(ϵ\right)$ where $g:{ℝ}^{+}\to {ℝ}^{+}$ is of our choosing, By picking $N=\max (N^{\prime },K)$ we know that

$$\begin{array}{cccc}& |b_n-L|& =|b_n-a_n+a_n-L|& \text{<span class="tml-eqn"></span>}\\ & & \le |b_n-a_n|+|a_n-L|& \text{<span class="tml-eqn"></span>}\\ & & \le \frac{1}{n}+f\left(ϵ\right)& \text{<span class="tml-eqn"></span>}\\ & & \le \frac{1}{K}+f\left(ϵ\right)& \text{<span class="tml-eqn"></span>}\\ & & \le g\left(ϵ\right)+f\left(ϵ\right)& \text{<span class="tml-eqn"></span>}\end{array}$$

to obtain the desired inequality one would have to pick $f,g$ such that $f\left(ϵ\right)+g\left(ϵ\right)<ϵ$, a simple choice of $f\left(x\right)=g\left(x\right)=\frac{x}{4}$ yields $|b_n-L|<\frac{ϵ}{2}<ϵ$

Let $a\in ℝ$ since $\left(x_n\right)$ converges then we know that there is some $N$ such that for all $n\ge N$ we have $|x_n-x|\le a$, therefore $|x_n|\le a+|x|$

Set $I:=\{0,...,N-1\}$ and we define $L:=\max \left(\{|x_i|:i\in I\}\right)$, then it should be clear that for any $k\in I$ we have $|x_k|\le L$

Finally, we can see that the value $M:=\max (L,a+|x|)$ has the property so that for any $j\in {\mathbb{N}}_0$ $|x_j|\le M$ so that $\left(x_n\right)$ is bounded.

Let $ϵ\in {ℝ}^{+}$ then there exists an $N\in {\mathbb{N}}_1$ such that $$|\sup \left(\{a_k:k\ge n\}\right)-L|<ϵ$$ which is true if and only if $$L-ϵ<\sup \left(\{a_k:k\ge n\}\right)<L+ϵ$$ thus since $a_n\le \sup \left(\{a_k:k\ge n\}\right)$ by transitivity we know that $a_n<L+ϵ$

Define $s_n:=\sup \left(\{a_k:k\ge n\}\right)$ and $i_n:=\inf \left(\{a_k:k\ge n\}\right)$.

Now suppose that $\mathrm{lim\hspace{0.17em}sup}\left(a_n\right)=\mathrm{lim\hspace{0.17em}inf}\left(a_n\right)=L$, since $a_k\in \{a_k:k\ge n\}$ , then as they are defined as upper and lower bounds, we have $i_n\le a_n\le s_n$ for all $n\in {\mathbb{N}}_1$ then by the squeeze theorem $\left(a_n\right)\to L$ .

Now suppose that $\mathrm{lim\hspace{0.17em}sup}\left(a_n\right)=\mathrm{lim\hspace{0.17em}sup}\left(a_n\right)=L$, and let's prove $\left(a_n\right)\to L$, let $ϵ\in {ℝ}^{+}$, therefore we get an $N\in {\mathbb{N}}_1$ such that for every $n\ge N$ we have $|a_n-L|<ϵ$ iff $L-ϵ\le a_n\le L+ϵ$, thus $L+ϵ$ and $L-ϵ$ are upper and lower bounds of $\{a_k:k\ge n\}$ respectively, since the sup and the inf are the lest and greatest of their bounds respectively so we have that $\sup \left(\{a_k:k\ge n\}\right)\le L+ϵ$ and $L-ϵ\le \inf \left(\{a_k:k\ge n\}\right)$ combining the two we obtain $$L-ϵ\le \inf \left(\{a_k:k\ge n\}\right)\le \sup \left(\{a_k:k\ge n\}\right)\le L+ϵ$$ which is equivalent to both $$|\inf \left(\{a_k:k\ge n\}\right)-L|\le ϵ\text{ and }|\sup \left(\{a_k:k\ge n\}\right)-L|\le ϵ$$ therefore $\mathrm{lim\hspace{0.17em}sup}\left(a_n\right)=\mathrm{lim\hspace{0.17em}inf}\left(a_n\right)=L$

TODO

Let $ϵ\in {ℝ}^{+}$, since $\left(x_n\right)$ is cauchy we obtain some $N_1$ such that for any $n,m\ge N,|a_n-a_m|\le \frac{ϵ}{2}$, on the other hand we see that there is some $N_2$ such that for all $n\ge N_2$ we have $|x_{\sigma \left(n\right)}-L|\le \frac{ϵ}{2}$, so let $N=\max (N_1,N_2)$ and let $n\ge N$, consider $\sigma \left(n\right)$, we will want this value to be greater or equal to $N_1$ so we can use it in-place of $m$, if it turns out that $\sigma \left(n\right)<N_1$ then since $\sigma$ is increasing there exists some $k>n\ge N_2$ such that $\sigma \left(k\right)\ge N_1$, then note: $$\begin{array}{cccc}& |x_n-L|& =|x_n-x_{\sigma \left(k\right)}+x_{\sigma \left(k\right)}-L|& \text{<span class="tml-eqn"></span>}\\ & & \le |x_n-x_{\sigma \left(k\right)}|+|x_{\sigma \left(k\right)}-L|& \text{<span class="tml-eqn"></span>}\\ & & <\frac{ϵ}{2}+\frac{ϵ}{2}=ϵ& \text{<span class="tml-eqn"></span>}\end{array}$$

We will construct a subsequence $\left(x_{\sigma \left(n\right)}\right)$ such that $|x_{\sigma }\left(k\right)-L|<\frac{1}{k}$. By doing so, we will have solved the problem, because for any $ϵ\in {ℝ}^{+}$ there is always a $k\in {\mathbb{N}}_1$ such that $\frac{1}{k}<ϵ$ which shows that $|x_{\sigma \left(k\right)}-L|<ϵ$ by transitivity.

Let $k\in {\mathbb{N}}_1$ since $\left(L_n\right)\to L$ then by considering $\frac{1}{2k}$ we know that there is an $N_k^{\prime }\in {\mathbb{N}}_1$ such that for all $n\ge N_k^{\prime }$ we have $|L_n-L|<\frac{1}{2k}$, now consider any $m_k\ge N_k^{\prime }$, thus we know that $|L_{m_k}-L|<\frac{1}{2k}$.

Now since $m_k\in {\mathbb{N}}_1$ then there exists a subsequence $x_{{\sigma }_{m_k}\left(n\right)}\to L_{m_k}$ , therefore using $ϵ=\frac{1}{2k}$ we get a $N_k$ such that for all $n\ge N_k,\left|x_{{\sigma }_{m_k}\left(n\right)-L_{m_k}}\right|<\frac{1}{2k}$, therefore we have

$$\begin{array}{cccc}& |x_{{\sigma }_{m_k}\left(n\right)}-L|& =|x_{{\sigma }_{m_k}\left(n\right)}-L_m+L_m-L|& \text{<span class="tml-eqn"></span>}\\ & & \le |x_{{\sigma }_{m_k}\left(n\right)}-L_{m_k}|+|L_{m_k}-L|& \text{<span class="tml-eqn"></span>}\\ & & <\frac{1}{2k}+\frac{1}{2k}=\frac{1}{k}& \text{<span class="tml-eqn"></span>}\end{array}$$

Initially it might seem as simple as taking the subsequence $x_{{\sigma }_{m_1}\left(N_1\right)},x_{{\sigma }_{m_2}\left(N_2\right)},x_{{\sigma }_{m_3}\left(N_3\right)},\dots$ but there's no guarentee that this subsequence has increasing indices (which is a requirement of a subsequence), therefore we need more work.

Condensing what we've proven above, we've shown that $$\forall k\in {\mathbb{N}}_1,\exists N_k\in {\mathbb{N}}_1,{\sigma }_{m_k}:{\mathbb{N}}_1\to {\mathbb{N}}_1\text{ st }\forall n\ge N_2,P(k,{\sigma }_m\left(n\right))$$

Where $P(k,v):=|x_v-L|<\frac{1}{k}$, and since ${\sigma }_{m_k}$ defines a subsequence then it is strictly increasing, therefore there exists an ordered sequence $({\alpha }_1\le {\alpha }_2\le {\alpha }_3\le \dots )$ such that $P(k,{\alpha }_k)$ which is to say that $|x_{{\alpha }_k}-L|<\frac{1}{k}$ so that the following subsequence works $$(x_{{\alpha }_1},x_{{\alpha }_2},\dots )$$ or in another notation with $\sigma \left(n\right)={\alpha }_n$ we are using the subsequence $\left(x_{\sigma \left(n\right)}\right)$

which is obtained by using the fact that we can insert $0=-a_i+a_i$ into the absolute value and then use the triangle inequality to break it, the general result is proven using induction using this fact in the induction step.

Let $ϵ\in {ℝ}^{+}$ and set $S_m:={\sum }_{i=1}^{k-1}|a_{i-1}-a_i|$ since we we assumed that ${\lim }_{N\to \infty }{S}_N$ exists, then the sequence is cauchy so we get some $N^{\prime }$ such that for any $a,b\ge N,|S_a-S_b|\le ϵ$ and take $N=N^{\prime }$ and let $n,m\ge N$, and without loss of generality assume that $n\le m$ $$\begin{array}{cccc}& |a_n-a_m|& \le |a_n-a_{n+1}|+|a_{n+1}-a_{n+2}|+\cdots +|a_{m-1}-a_m|& \text{<span class="tml-eqn"></span>}\\ & & =S_m-S_n& \text{<span class="tml-eqn"></span>}\\ & & =|S_m-S_n|\le ϵ& \text{<span class="tml-eqn"></span>}\end{array}$$ where we've used the generalized triangle inequality and the fact that $S_k$ is increasing, as each higher index adds a non-negative element to the sum.

Recall that a geometric series of halves is bounded, and thus we're inspired to try to make each $|x_{\sigma \left(k\right)}-x_{\sigma (k+1)}|\le \frac{1}{2^k}$, call this property $\alpha$.

Let's construct this subsequence, for any $k\in {\mathbb{N}}_1$ consider $ϵ=\frac{1}{2^k}$, then we get an $N_k$ such that for all $n,m\ge N_k,|a_n-a_m|\le \frac{1}{2^k}$, and consider the sequence $x_{N_1},x_{N_2},x_{N_3},\dots$, we'll show it satisfies $\alpha$.

Let $j\in {\mathbb{N}}_1$, and consider $N_j,N_{j+1}$, in the case that $N_j\ge N_{j+1}$ then note that $|x_{N_j}-x_{N_{j+1}}|\le \frac{1}{2^{j+1}}\le \frac{1}{2^j}$ (where $n=N_j,m=N_{j+1}$), in the case that $N_j<N_{j+1}$ we also get that $|x_{N_j}-x_{N_{j+1}}|\le \frac{1}{2^j}$ so $\alpha$ is satisfied for this subsequence. Therefore $$\sum _{k=1}^{\infty }|x_{N_k}-x_{N_{k+1}}|\le \sum _{k=1}^{\infty }\frac{1}{2^k}=1$$ as needed.

$⟹$ this direction is easy, let $ϵ\in {ℝ}^{+}$, since $\left(a_n\right)$ is cauchy we get some $N^{\prime }$ such that for all $n,m\ge N^{\prime },|a_n-a_m|<ϵ$, and so take $N=N^{\prime }$ then specifically for $n,n+1\ge N$ we have that $ϵ>|a_n-a_{n+1}|=||a_n-a_{n+1}|-0|$ as needed.

$⟸$ Assume that ${\lim }_{n\to \infty }|a_n-a_{n+1}|=0$ and now we want to prove that $\left(a_n\right)$ is cauchy so let $ϵ\in {ℝ}^{+}$ and so we obtain some $N$ such that for any $n\ge N$ we have that $|a_n-a_{n+1}|\le ϵ$ take $N^{\prime }=2N+1$ and let $n,m\ge N^{\prime }$ by our lemma, we see that $a_n,a_m\in [a_{2N},a_{2N+1}]$, but then again since $2N,2N+1\ge N$ we see that $|a_{2N+1}-a_{2N}|\le ϵ$ and therefore $|a_n-a_m|\le ϵ$ as needed.