I'm struggling to show that the sum $$\sum_{i=1}^{\infty} \left(A_i+\frac{2\varepsilon}{2^i}\right)^s$$ can be taken as $\sum_{i=1}^{\infty}\left(A_i\right)^s$ for arbitrary $\varepsilon>0$. So it's showing that the fraction added can be taken to just be $0$.
Also, I'm new so apologies if things aren't quite right. I was struggling to word the question as well.
First of all, Welcome to the site !
Consider $$a_i=\left(A_i+2^{1-i} \epsilon \right)^s\implies \log(a_i)=s \log\left(A_i+2^{1-i} \epsilon \right)$$ Use Taylor series to get $$\log\left(A_i+2^{1-i} \epsilon \right)=\log (A_i)+\frac{2^{1-i} \epsilon }{A_i}-\frac{2^{1-2 i} \epsilon ^2}{A_i^2}+O\left(\epsilon^3\right)$$ Multiply by $s$ and continue with Taylor since $$a_i=e^{\log(a_i)}=A_i^s+2^{1-i} s \epsilon A_i^{s-1}+2^{1-2 i} (s-1) s \epsilon ^2 A_i^{s-2}+O\left(\epsilon^3\right)$$ Now, make $\epsilon$ very small and make the summation.