Show that every subsequence converging to the same limit implies limit of sequence exists. I need to prove that if $\exists$ subsequences $a_{n_{k_{1}}}$ and $a_{n_{k_{2}}}$ (of $a_{n}$) that converge to different limits, then the sequence $a_{n}$ does not converge. I'm not sure how to do this. If I suppose $a_{n_{k_{1}}}$ converges to some limit $l_{1}$, then $\forall \epsilon >0$, $\exists n_{k_{1}}>N$ s.t. $|a_{n_{k_{2}}}-l_{1}|<\epsilon$, and the same thing for $a_{n_{k_{2}}}$ and its limit $l_{2}$. 
$a_{n}$ not converging means $\exists \epsilon_{0}$ s.t. $\forall N\in \mathbb{N}$, $\exists n_{k}\geq N$ s.t. $|a_{n}-l|\geq \epsilon_{0}$. 
But, I'm not sure how to put them together in order to prove the implication that I want.

Suppose $a_n$ does converge to some value $\alpha$.
This means that $\forall \epsilon>0 \exists N>0$ s.t $\forall n >N$: $|a_n - \alpha| < \epsilon$.
Let $a_{n_k}$ be a subsequence of our original $a_n$. Since both sequences are infinite, and $N$ is a finite number, we can infer that there is some $\beta$ such that $\forall i > \beta$: $a_{n_i} \in \{a_n,a_{n+1},...\}$ where $n>N$.
we already know that every element in the set $\{a_n,a_{n+1},...\}$ agrees with $|a_i-\alpha|<\epsilon$, and that $a_{n_i} \in \{a_n,a_{n+1},...\}$, so in particular, $|a_{n_i} - \alpha| < \epsilon$.
So overall: for all $\epsilon >0 \exists \beta>0$ s.t $\forall i>\beta$: $|a_{n_i} - \alpha|<\epsilon$, so $\alpha$ is the limit of $a_{n_k}$.
Result: if $a_n$ converges to $\alpha$, then ANY infinite subsequence of $a_n$ also converges to $\alpha$.
So if two subsequences are converging to different limits, then that means the entire sequence does not converge.
