Whe had the following theorem in class:
If $(a_{2n})_{n\in\mathbb N}$ and $(a_{2n+1})_{n\in\mathbb N}$ are convergent sequences with the same limit $a$, then the sequence $(a_{n})_{n\in\mathbb N}$ converges.
Proof:
$(a_{2n})_{n\in\mathbb N}$ converges, so $\forall\varepsilon>0\exists N_1\in\mathbb N:\forall n\geq N_1: |a_{2n}-a|<\varepsilon$
It follows $|a_m-a|<\varepsilon$ for all even $m\geq2N_1$.
$(a_{2n+1})_{n\in\mathbb N}$ converges, so $\forall\varepsilon>0\exists N_2\in\mathbb N:\forall n\geq N_2: |a_{2n+1}-a|<\varepsilon$
It follows $|a_m-a|<\varepsilon$ for all odd $m\geq2N_2+1$.
Thus for all $m\in\mathbb N$ with $m\geq M:=\max\{2N_1,2N_2+1\}$ is $|a_m-a|<\varepsilon$ $\;\;\;\;\;\;\;\;\;\;\;$ q.e.d.
Now, after trying to proove it myself again, I've just wondered myself about one thing:
Why can't you choose $M=\max\{N_1,N_2\}$ and leave the second line in each number out? Why do you have to choose $2N_1$ and $2N_2+1$ ?
Thanks for helping!