Proving a sequence with infinitely many zeros is not zero heavy?

Question

Is my counterexample correct?

If a sequence contains an infinite number of zeroes, is it necessarily zero-heavy? If not, provide a counterexample.

Solution. Consider the sequence $(a_n)$ defined such that $a_n = 0$ whenever $n = 2^k$ for some $k\in\mathbf{N}$ and is $1$ otherwise, then given any $M\in\mathbf{N}$ we may choose a $k\in\mathbf{N}$ such that $2^{k+1}-(2^{k}+1)>M$ then given the construction of the sequence if $n\in\{2^{k}+1,2^{k+1}+2,\dots,(2^{k}+1)+M\}$ we have $a_n\neq 0$.The sequence in question then cannot possibly be zero heavy.

NOTE: We define a sequence to be zero heavy if $$\exists M\in\mathbf{N}\forall N\in\mathbf{N}\exists n\in\{N,N+1,\dots,N+M\}(x_n=0)$$

Answer 1

Looks good! The only part I might do/phrase at all differently is:

...we may choose $k\in\mathbf{N}$ such that $k>\log_2(M+1),$ so that $2^{k+1}-(2^k+1)>M,$ and so, given the construction of the sequence, if....