Let $A$ be some unital Banach *-algebra. M. Takesaki, Theory of Operator Algebras, vol. 1. Springer, 1979. claims that if $|\lambda|>r(x):=\lim\inf_{n\rightarrow\infty}\|x^n\|^{1/n}$ then $\sum_{n=0}^\infty x^n/\lambda^{n+1}$ converges. I don't see why this is the case. This is used to claim that $x-\lambda$ is invertible. I would appreciate if someone could help me pin point why the sum converges.
As an alternative, the way that W. Rudin, Functional Analysis, 2nd ed. McGraw-Hill, Inc., 1991. does this is by showing that if $x-\lambda$ is not invertible then $|\lambda|\leq \inf_{n\geq 1}\|x^n\|^{1/n}\leq r(x)$. For this he notes that $x^n-\lambda^n=(x-\lambda)(x^{n-1}+x^{n-2}\lambda+\cdots\lambda^{n-1})$. He claims that from this decomposition it is clear that $x^n-\lambda^n$ can't be invertible. Once again, I don't see why this is true, for a product containing a non-invertible element can still be invertible.
Mark solved the second half in the comments. As far as the first half is concerned, I saw the answer in O. Bratteli and D. W. Robinson, Operator Algebras and Quantum Statistical Mechanics, Second., vol. 1. Springer, 1987. The idea is that $|\lambda|>r(x)$ implies that there is an $n$ such that $|\lambda^n|>\|x\|^n$. Now, due to the Euclidean algorithm, one can write $$\sum_{n=0}^\infty\frac{x^n}{\lambda^{n+1}}=\frac{1}{\lambda}\sum_{p=0}^\infty\sum_{q=0}^{n-1}\left(\frac{x}{\lambda}\right)^{np+q}=\frac{1}{\lambda}\sum_{q=0}^{n-1}\left(\frac{x}{\lambda}\right)^{q}\sum_{p=0}^{\infty}\left(\frac{x^n}{\lambda^n}\right)^{p}.$$ In this last form, convergence is clear.