Convergence of sequence (write a proof) I need to prove the following affirmation: If $ \lim x_{2n} = a $ and $ \lim x_{2n-1} = a $, prove that $\lim x_n = a $ (in $ \mathbb{R} $ )
It is a simple proof but I am having problems how to write it. I'm not sure it is the right way to write, for example, that the limit of $(x_{2n})$ converges to a:
$ \forall \epsilon > 0 \; \exists n_1 \in \mathbb{N} $ such that if $n \in \mathbb{N}$ and $n \geq n_1$, then (*) $ | x_{2n} - a| < \epsilon $
About the (*) step, is that correct to write $ x_{2n} $? Or  should I use another notation?
Thanks for the help! 
 A: You can do so. As a matter of fact,you consider the sequence $(y_n)$ given by $y_n=x_{2n}$ and could write - as you may be more accustomed to - that $|y_n-a|<\epsilon$ for all $n>n_1$. But as $y_n=x_{2n}$ you really get back what you (correctly)  wrote: $|x_{2n}-a|<\epsilon$.
A: $$\forall\varepsilon>0 \; \exists n_1\in\mathbb{N} \; \forall k>n_1 |a_{2k} - a|<\varepsilon
\\ \forall\varepsilon>0 \; \exists n_2\in\mathbb{N} \; \forall k>n_2 |a_{2k+1} - a|<\varepsilon$$
Therefore if we set $n'=\max\{n_1,n_2\}$ we get
$$\forall\varepsilon>0 \; \exists n'\in\mathbb{N} \; \forall k>n' \;|a_{2k} - a|<\varepsilon
\\ \forall\varepsilon>0 \; \exists n'\in\mathbb{N} \; \forall k>n' \; |a_{2k+1} - a|<\varepsilon$$
As $2k>2n', 2k+1>2n'+1$ we can set $n=2n'$ and get
$$\forall\varepsilon>0 \; \exists n\in\mathbb{N} \; \forall k>n \;|a_k - a|<\varepsilon$$
A: Hint: for $\epsilon>0$ find some $n_{1}$ such that for $n>n_{1}$ we have $|x_{2n}-a|<\epsilon$ and also some $n_{2}$ such that for $n>n_{2}$ we have $|x_{2n-1}-a|<\epsilon$. Based on that try to find some $n_{0}$ such that  for $n>n_{0}$ we have $|x_{n}-a|<\epsilon$
