Prove that if $a_{2n} \rightarrow g$ and $a_{2n+1} \rightarrow g$ then $a_n \rightarrow g$ The problem is in the question:
Prove that if sequences $a_{2n} \rightarrow g$ and $a_{2n+1} \rightarrow g$ then $a_n \rightarrow g$.
I don't know how to prove that - it seems obvious when we look at the definition that for sufficiently large $n$ (let's say $n>N$) we have $|a_{2n}-g|$ and $|a_{2n+1}-g|$ less than any given $\epsilon$ and these  are all elements of $a_n$ when $n>N$ (even and odd terms). Does this need more formal proof?

Yes, for a given $\varepsilon>0$, you will get an $N_1$ from $(a_{2n})$ and an $N_2$ from $(a_{2n+1})$ and you will need to take their maximum or so..
