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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Initially it might seem as simple as taking the subsequence $$ x_{{\mathrm{\sigma <!--\; \sigma \; -->}}_{m_1}\left(N_1\right)},x_{{\mathrm{\sigma <!--\; \sigma \; -->}}_{m_2}\left(N_2\right)},x_{{\mathrm{\sigma <!--\; \sigma \; -->}}_{m_3}\left(N_3\right)},\dots <!--\; \dots \; -->$$ 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 $\[ \mathrm{\forall <!--\; \forall \; -->}k\in <!--\; \in \; -->{\mathbb{N}}_1,\mathrm{\exists <!--\; \exists \; -->}{N}_k\in <!--\; \in \; -->{\mathbb{N}}_1,{\mathrm{\sigma <!--\; \sigma \; -->}}_{m_k}:{\mathbb{N}}_1\to <!--\; \to \; -->{\mathbb{N}}_1\text{ st }\mathrm{\forall <!--\; \forall \; -->}n\ge <!--\; \ge \; -->N_2,P(k,{\mathrm{\sigma <!--\; \sigma \; -->}}_m\left(n\right)) \]$

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

Let $$ \mathrm{ϵ<!--\; ϵ\; -->}\in <!--\; \in \; -->{\mathbb{R}}^{+}$$ and set $$ S_m:=\underset{i=1}{\overset{k-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}|<!--\; |\; -->a_{i-<!--\; -\; -->1}-<!--\; -\; -->a_i|<!--\; |\; -->$$ since we we assumed that $$ \underset{N\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{S}_N$$ exists, then the sequence is cauchy so we get some $$ N^{\mathrm{\prime <!--\; \prime \; -->}}$$ such that for any $$ a,b\ge <!--\; \ge \; -->N,|<!--\; |\; -->S_a-<!--\; -\; -->S_b|<!--\; |\; -->\le <!--\; \le \; -->\mathrm{ϵ<!--\; ϵ\; -->}$$ and take $$ N=N^{\mathrm{\prime <!--\; \prime \; -->}}$$ and let $$ n,m\ge <!--\; \ge \; -->N$$, and without loss of generality assume that $$ n\le <!--\; \le \; -->m$$ 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 $$ |<!--\; |\; -->x_{\mathrm{\sigma <!--\; \sigma \; -->}\left(k\right)}-<!--\; -\; -->x_{\mathrm{\sigma <!--\; \sigma \; -->}(k+1)}|<!--\; |\; -->\le <!--\; \le \; -->\frac{1}{2^k}$$, call this property $$ \mathrm{\alpha <!--\; \alpha \; -->}$$.

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

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

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

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