For $k\geq3$, the equation $f(x)=\frac{1}{2}(1+x^{k+1})$ has a fixed point $x_k$ satisfying $\frac{1}{2} < x_k < \frac{3}{5}$. Since $f$ is increasing, we can apply it repeatedly to this pair of inequalities, to obtain new upper and lower bounds for $x_k$. I'm told it follows that $x_k = \frac{1}{2} + \frac{1}{2^{k+2}} + O \left ( \frac{k}{2^{2k}} \right )$, but I can't see why.
So far I have tried this: Apply $f$ twice to the upper bound, then apply the binomial theorem, to obtain a new upper bound of $\frac{1}{2} + \frac{1}{2^{k+2}} \left (1 + \sum_{j=1}^{k+1} \binom{k+1}{j} \left ( \frac{3}{5} \right )^{j(k+1)} \right )$. Then I tried to bound the binomial coefficients above by the middle one, and use Sterling's approximation, but I couldn't get what I needed.
So, by definition, $1/2 < x_k < 3/5$ with $2x_k=1+x_k^{k+1}$. So $1 \leq 2x_k \leq 1+(3/5)^{k+1}$, so $x_k \rightarrow 1/2$. Write $x_k=1/2+y_k$, $y_k \geq 0$, $y_k \leq 0.5\cdot (3/5)^{k+1}$, so $ky_k \rightarrow 0$.
Then $2y_k=(1/2)^{k+1}(1+2y_k)^{k+1} \geq 2^{-k-1}$, and $2y_k \leq (1/2)^{k+1}(e^{2y_k})^{k+1}\leq (1/2)^{k+1}\exp{(k+1)0.6^{k+1}} \leq \frac{1}{2^{k+1}}+(e-1)(k+1)0.3^{k+1}$ as long as $(k+1)0.6^{k+1} < 1$, eg when $k > 3$.
So $0 \leq r_k:=y_k-\frac{1}{2^{k+2}} \leq (e-1)(k+1)0.3^{k+1} = o(0.4^{-n})$ thus $y_k \sim 2^{-k-2}$.
We know that $2^{k+2}y_k \leq e^{2y_k(k+1)} = 1+O(y_k(k+1))$.
We can rewrite this as $2^{k+2}r_k \leq O(y_kk)=O(k2^{-k})$, therefore $0 \leq r_k \leq O(k2^{-2k})$, QED.