Suppose we have the Hilbert entropy function with $0\leq\lambda\leq\frac{1}{2}$ $$H(\lambda) = \lambda \ln{\frac{1}{\lambda}+(1-\lambda)\ln{\frac{1}{1- \lambda}}}.$$
I would like to show using Stirling's formula applied to the natural logarithm $$\ln{n!} = n \ln{n} - n + O(\ln{n})$$ that $$n^{-1}ln {\binom{n}{k}} = H(\lambda ) + o(1)$$ with $k = \lambda n + O(1)$.
I tried the following steps: $$n^{-1}(\ln{n!} - \ln{k!} - \ln{(n-k)!})=$$ $$n^{-1}(n\ln{n}-n-k\ln{k}-k-(n-k)\ln(n-k)-(n-k)+O(\ln{n}))=$$ $$n^{-1}(n\ln{n}-2n-k\ln{k}-(n-k)\ln{(n-k)}+O(\ln{n}))$$. After resubstituting $\lambda n$ I get: $$n^{-1}(n\ln{n}-2n-(\lambda n)\ln{\lambda n}-(n-\lambda n)\ln{(n-\lambda n)}+O(\ln{n}))=$$ $$n^{-1}(n\ln{n}-2n-(\lambda n)\ln{\lambda n}-(1-\lambda)n\ln{(1-\lambda)n}+O(\ln{n}))= \dots$$ And from there I really don't know what to do next. The coefficients are somewhat similar to those of the entropy function but the inside is not a fraction. When using the $\ln{(a^b)}=b \ln(a)$ property with $-1$ I get a nonsense result. Have I made mistakes in applying Stirling's formula?