If $\{u_{k}\}_{k=1}^{\infty}$ is a sequence in an Banach algebra (and more specifically, in the set of all the bounded linear operators of a Banach space $X$). If $$\sum_{k=1}^{\infty}\lambda^{k}u_{k}=0,\quad \text{for every $\lambda\in \mathbb{D}(0, 1/3)$},$$ where the $\mathbb{D}(0, 1/3)~$ denotes an open disc centered at $0$ and radius $1/3$.
Then can we conclude that $\boldsymbol{ u_{n}=0}$, for every $\boldsymbol{n}$?
The answer is yes.
The fact that $\sum_{k=1}^\infty \lambda^k u_k=0$, for all $\lambda\in \mathbb D(0,1/3)$, implies that the sequence $s_n=\sum_{k=1}^n \lambda^k u_k\in {\mathscr B}(X)$, $n\in\mathbb N,$ converges, for all $\lambda$, which implies that $\lambda^n u_n\to 0$, for every $|\lambda|<1/3$, thus $\limsup_{n\to\infty}|\lambda|\|u_n\|^{1/n}\le 1 $, for every $|\lambda|<1/3$, and hence $\limsup_{n\to\infty}\|u_n\|^{1/n}\le 3,$ which in turn implies that the function $$ f(\lambda)= \sum_{k=1}^\infty \lambda^k u_k, \quad \lambda\in\mathbb D(0,1/3), $$ is in $C^\infty\big(\mathbb D(0,1/3);\mathscr B(X)\big)$. Since $f(\lambda)=0$, for every $\lambda\in \mathbb D(0,1/3)$, then $f^{(n)}(\lambda)=0$, for all $n$ and $\lambda\in\mathbb D(0,1/3)$. In particular $0=f^{(0)}(0)=u_n.\qquad$ QED