Are there any techniques that can be used in the case when a Neumann series doesn't converge?

Question

Suppose we have a bounded linear operator $A = A(\gamma):H_1\to H_2$ where $H_1$ and $H_2$ are Hilbert spaces and $\gamma>0$ is some parameter, and we are interested in the solution to $$ (I-A)x = y. $$ If $\|A\|<1$ we can use a Neumann series expansion and get a series representation: \begin{align} x & = (I-A)^{-1} y \\ & = \sum_{j=0}^\infty A^j y \end{align}

Now, suppose that when $\gamma$ becomes small enough $\|A(\gamma)\|$ becomes greater than $1$ and the Neumann series won't converge; thus we can't get a series representation for $x$.

An example of this situation can be found in problems featuring scattering of waves among disjoint objects. If the solution is represented as a Neumann series, it can fail to converge if frequencies become high or if distances between objects become small.

Are any techniques that can used in such a situation to 'get around' the non-convergence of a Neumann series and obtain a series representation for $x$?

Answer 1

If the spectrum of $A$ is contained in a disk $\{z: |z - a| \le r\}$ where $|1-a| > r$, then the series $\sum_{n=0}^\infty (1-a)^{-1-n} (A - a I)^n$ converges to $(I-A)^{-1}$.

Answer 2

Of course, one only has a chance if $1$ is not in the spectrum of $A$.

Robert Israel's answer gives a series that converges to the resolvent $(I-A)^{-1}$ if the spectrum of $A$ is, for instance, contained in a disk with radius larger then $1$, but centered sufficiently far in the left half plane.

Another method to obtain a series representation for $(I-A)^{-1}$ is based on the following fact:

Proposition. Let $\mu \in \mathbb{C}$ be in the resolvent set of $A$ and let $\lambda \in \mathbb{C}$ be a number such that $|\lambda - \mu| < \frac{1}{\|(\mu I - A)^{-1}\|}$. Then $\lambda$ is also in the resolvent set of $A$ and the resolvent of $A$ at $\lambda$ is given by $$ (\lambda I - A)^{-1} = \sum_{k=0}^\infty (\mu - \lambda)^k (\mu I - A)^{-(k+1)}. $$ Proof. This follows readily from the Neumann series expansion if we use that $$ \lambda I - A = (\lambda - \mu) I + \mu I - A = (\mu I - A)^{-1}\Big( I - (\mu - \lambda)(\mu I - A) \Big). $$

If one is interested in the case $\lambda = 1$, but the spectral radius of $A$ is $\ge 1$, one could for instance try the following procedure:

Choose a real number $r > r(A)$ (where $r(A)$ denotes the spectral radius) and use the Neumann series to compute $(rI - A)^{-1}$. In case that the interval $[1,r]$ is in the resolvent set of $A$, one can now move a bit left of $r$ and compute the resolvent at this new point by means of the above proposition. Then, again, one can move a bit more left, and iterate this procedure until one arrives at $1$. Thus, one obtains a "representation" of $(I - A)^{-1}$ by means of a finitely often iterated series expansion.

Whether this is useful or not depends of course on the application one has in mind. Sometimes this (or a related) technique can be quite useful for theoretical purposes; on the other hand, I would suspect that the procedure is completely unsuited for, say, numerical computations.