For a given $k \in \mathbb{N}$, could you determine $f_k(n)$ such that the following holds?
$$\sum_{i=0}^n \left (1-\frac{1}{2^i} \right )^k - f_k(n) \rightarrow 0$$
For $k=1$,
$$ \sum_{i=0}^n \left (1-\frac{1}{2^i} \right )=n-2.$$
Hence, $f_1(n)=n-2$.
We have $$\sum_{i=0}^n(1-2^{-i})^k=\sum_{i=1}^n(1-2^{-i})^k=\sum_{i=1}^n\left [1+\sum_{l=1}^k{k\choose l}(-1)^l2^{-il}\right ]\\ =n+\sum_{l=1}^k{k\choose l}(-1)^l\sum_{i=1}^n2^{-il}=n+\sum_{l=1}^k{k\choose l}(-1)^l{2^{-l}\,[1-2^{-nl}]\over 1- 2^{-l}}$$ Hence $$f_k(n)=n+\sum_{l=1}^k{k\choose l}(-1)^l{1\over 2^l- 1}$$ In particular $f_1(n)=n-1.$
Another solution By the limit test the following series is convergent $$\sum_{i=1}^\infty [1-(1-2^{-i})^k]$$ as $$\lim_{i\to \infty}{1-(1-2^{-i})^k\over 2^{-i}}\underset{x:=1-2^{-i}}{\quad =\quad }\lim_{x\to 1}{1-x^k\over 1-x}=k$$ Hence $$ \sum_{i=1}^n(1-2^{-i})^k-n=-\sum_{i=1}^n[1-(1-2^{-i})^k]\underset{n\to \infty}{\longrightarrow} -\sum_{i=1}^\infty [1-(1-2^{-i})^k]$$ Thus $$f_k(n)=n-\sum_{i=1}^\infty [1-(1-2^{-i})^k]$$