Why does {0,1,0,0,1,0,0,0,1,0,0,0,0,1...} diverge? My book says that this sequence diverges because it "takes on only two values, 0 and 1, and never stays arbitrarily close to either one for n sufficiently large." However, I don't understand why this cannot converge. My intuition says that for an infinite n, we will reach a point where there are infinitely many 0's before hitting a 1, because each repetition, we get more and more zeroes. Eventually, wouldn't we have infinite zeros, and the pattern would converge to 0?
 A: 
Eventually we will have infinite zeros

No, you will not. The number of zeros between each pair of ones gets bigger and bigger, but it is always finite. Consider that each term in the sequence only has a finite number of terms that come before it. So if after some instance of 1, you have infinitely many zeros, how many zeros precede this 1?
Moreover, we have a definition for a real sequence converging:

A sequence $a_n$ converges to a point $a$ if for every $\epsilon>0$, there exists $N\in\mathbb N$ such that if $n\geq N$, then $|a_n-a|<\epsilon$.

So if $\epsilon=1/2$, how big must $N$ be?
A: Consider the definition of convergence for a sequence. For any positive $\varepsilon$, you want to find a $M \in \mathbb N$  such that whenever $n \ge M$, then  $\left| x_n \right| < \varepsilon$.
However, for $0 \le \varepsilon < 1$, you wont be able to find such an $M$, since at some point, there will always be $X_n = 1$, even though $n \ge M$
A: Remember the epsilon-delta definition of the limit. To prove that this converges, you need to give me what the limit is, and the neighborhood of x I can choose from. If I cannot find an x that exceeds your limit, then the sequence is convergent.
So let's say you want to show that the sequence converges to any value less than 1, and the neighborhood I can choose from is a very large number to infinity. No matter how  far we go, there will ALWAYS be a 1 I can choose, which means the limit cannot be less than 1.
A: If you claim this sequence (call its terms $x_n$) converges to $0$, then you must provide some finite $N$ such that $x_n = 0$ for all $n \geq N$. There can be no such $N$, however, since $1$ appears an infinite number of times in the sequence.
