Does this sequence converge to $0$? In a similar vain as my previous question, I was wondering whether the following converges to $0:$
$$
a_n = \left\{ 
  \begin{array}{l l}
    1 & \quad n=2^k\,\text{for some}\,k\in\mathbb{N}_0\\
    0 & \quad \text{otherwise}
  \end{array} \right.$$
First terms are: 
$$1,1,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0\cdots$$
In other words, the $1$'s get more and more separated. Again, as in the other question, intuitively it would seem that "at infinity", the string of $0$'s is infinite, but since intuition can often be erroneous in cases like these, I am not convinced. Can someone enlighten me here?
Follow-up:
Can I conclude from the commments and answers that a sequence of integers that converges to $0$ must be $0$ for $n$ sufficiently large?
 A: For $\epsilon=\frac12$, there is no $N\in \mathbb N$ with the property that for all indices $n>N$ we have $|a_n|<\epsilon$. 
Even though they are somewhat rare, terms $a_n$ with $a_n=1>\epsilon$ occur again and again. (To be explicit, we can consider $n=2^N$, which is $>N$ and gives us $a_n=1$).
There is something behind this: The sequence indeed converges to zero in a less strict sense so to speak, see Cesáro mean.
A: The other answers are good, this is more of a side-topic or another place to look.
There is a notion of something called the "Cesàro mean." Given a convergent sequence, the Cesàro mean is just the limit. But the Cesàro mean also gives a value for some non-convergent sequences.  
For example, $-1,1,-1,1,\dots$ has Cesàro mean zero, but does not have a limit.
Your series has Cesàro mean zero, too.
A: No.  A sequence converges to zero if and only if we can eventually make all the terms which are large enough as close to zero as we want.  In symbols, $a_n \to a$ if for all $\epsilon > 0$, there exists a constant $M$ so that $n > M$ implies $|a_n - a| < \epsilon$. 
This cannot happen with your example, since every entry that is a power of two will be $1$.
