The Gilbert - Varshamov bound states for the binary case: $$ {\frac {2^{n}}{2^{k}}} \le \sum _{{j=0}}^{{d-1}}{\binom {n-1}{j}} $$
In the Nielsen - Chuang textbook (p 449) it is given (for large $n$) as: $$ \frac{k}{n}\geq{1-H(\frac{2t}{n})} $$
In the above $H(x)$ is the binary Shannon entropy, $d$ is the distance of the code and $t$ are the bits in error. What are the steps to go from the first description to the second?
There is a known bound
$$ \sum_{k=0}^m \binom{n}{k} \le 2^{nH(m/n)} \tag{1}$$
Taking logarithms in the Gilbert - Varshamov bound, we get
$$n-k = \log_2\left({\sum _{{j=0}}^{{d-1}}{\binom {n-1}{j}}}\right)\le n H\left(\frac{d-1}{n-1}\right)) = nH\left(\frac{2t}{n}\right)$$
where I've used $d = 2t+1$ and $ n \to \infty$.