Let's consider functional $$f: (c, \Vert \cdot\Vert_\infty)\ni (x_n) \rightarrow \sum_{n=1}^\infty \frac{x_n-\lim_{n \rightarrow\infty} x_n}{2^n}$$
where $c$ is a space of convergent sequences.
I want to calculate $\Vert f \Vert$.
My work so far
Let's denote $\lim_{n \rightarrow \infty }x_n = g$, then
$$f(x_n) = \frac{x_1-g}{2} + \frac{x_2-g}{2^2} + \frac{x_3-g}{2^3} + ... \le \frac{\Vert x_n \Vert_\infty - g}{2}+ \frac{\Vert x_n \Vert_\infty - g}{4}+ \frac{\Vert x_n \Vert_\infty-g}{8} + ...$$ $$= \Vert x_n\Vert_\infty-g$$
So when it comes to calculate norm I have for $\Vert x_n \Vert_\infty\le 1$:
$$|f(x_n)| \le 1-g$$
So I can see that supremum is hold for such sequnece $x_n$ such we have $|f(x_n)| = 1-g$.
Example of a sequence which satisfies it is $x_n = \frac 1 n$ ( $\Vert x_n \Vert_\infty = 1, \; g =0$)
But! I want to have as big value of $1-g$ as possible for sequence $x_n: \sup_n|x_n| \le 1$
If I'm correct then I'm looking for a sequence $x_n$ which statisfies the supremum condition and maximazes $1 - \lim_{n \rightarrow \infty}x_n$.
Could you please tell me how this sequnece can be chosen, and how can I prove that it really maximazes $1-g$ ?