# Is the Neumann series a compact operator?

Let $X$ be an infinite dimensional Banach space and $A:X\to X$ be a compact operator with the operator norm $\|A\|<1$. Then $I-A$ is invertible and the Neumann series $$S_N = \sum_{k=0}^N A^k$$ converges in the operator norm to $(I-A)^{-1}$: $$\|S_N-(I-A)^{-1}\| \to 0, \ \text{ as } \ N\to \infty$$ Now I think all $S_N$ are compact operators hence the limit $(I-A)^{-1}$ is also compact. However this cannot be the case because then $$I=(I-A)(I-A)^{-1}$$ would be compact, which is not possible for infinite dimension $X$. What is wrong in my argument?

• $A^0 = I$ is not compact (except in trivial cases). Since $A^n$ is compact for $n > 0$, none of the $S_N$ is compact (they are all compact perturbations of the identity). – Daniel Fischer Apr 22 '14 at 13:22

As pointed out by Daniel Fischer, it is true that for each $n\geqslant 1$, $A^n$ is compact hence so is $\sum_{n=1}^NA^n$. Since a norm-limit of compact operators is compact and $\sum_{n=1}^\infty\lVert A^n\rVert$ is convergent, we obtain that $\color{red}{A}(I-A)^{-1}$ is compact.
But $(I-A)^{-1}$ is not compact, otherwise so would be $A(I-A)^{-1}-(I-A)^{-1}=-I$.