Proving a Sequence Does Not Converge I have a sequence as such:

$$\left( \frac{1+(-1)^k}{2}\right)_{k \in \mathbb{N}}$$

Obviously it doesn't converge, because it alternates between $0,1$ for all $k$. But how do I prove this fact?
More generally, how do I prove that a sequence does not converge? Are there any neat ways other than "suppose, for contradiction, that the sequence converges. Then..."?
 A: In this case (but not universally), it suffices to note that the difference between two successive terms does not converge to $0$. 
Which begs the question of the reason why a sequence $(a_n)$ such that $|a_n|=1$ for every $n$, does not converge to $0$...
A: A sequence of real numbers converges if and only if it's a Cauchy sequence; that is, if for all $\epsilon > 0$, there exists an $N$ such that
$$n, m \ge N \implies |a_n - a_m| < \epsilon$$
If you haven't shown or seen this, I'd strongly suggest trying to prove it (or look it up in pretty much any basic analysis book).  
No matter what we choose for $N$ here, however, just choose $n = N + 1$ and $m = N$, with $\epsilon = \frac{1}{2}$. Then $|a_n - a_m| = 1 > \epsilon$, a contradiction.

Alternatively, show that if a sequence is convergent, then every subsequence is convergent to the same limit. Then choose appropriate subsequences.
A: $\{a_k\}$ converges if and only if $\{a_{2k}\}$ and $\{a_{2k+1}\}$ converge to the same limit. Here
$$
a_{2k}=1\rightarrow1, \quad a_{2k+1}=0\rightarrow 0.
$$
