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  "body": "\n## Statement\n\nLet $X$ be a [[Banach space|06GZ956657]], and let $T \\colon X \\to X$ be a [[continuous linear map|03TM135G0N]] with [[operator norm|06EKW1AKXF]] $\\norm{T} \u003c 1$.\nWrite $I$ for the identity map of $X$ and $T^0 = I$.\nThen the following hold.\n\n1. The series $\\sum_{n=0}^{\\infty} T^n$ converges in operator norm.\n2. The operator $I - T$ is [[invertible|CPAD5W1CGW]], and\n$$(I - T)^{-1} = \\sum_{n=0}^{\\infty} T^n.$$\n3. $\\norm{(I - T)^{-1}} \\le \\dfrac{1}{1 - \\norm{T}}$.\n\n## Remarks\n\nProof idea: $\\norm{T^n} \\le \\norm{T}^n$, so the series is dominated by a [[geometric series|01X5CVV1Q5]] and converges absolutely in the Banach space of bounded operators on $X$.\nSee [[absolutely convergent series in a Banach space converge|082MGNWNA8]].\nThe partial sums $S_N = \\sum_{n=0}^{N} T^n$ satisfy $(I - T) S_N = S_N (I - T) = I - T^{N+1}$, and $T^{N+1} \\to 0$.\nLetting $N \\to \\infty$ gives both inverse identities, and the norm bound follows from the geometric series.\n\nExample: on $\\ell^2$, let $T(x_1, x_2, \\ldots) = (0, \\tfrac12 x_1, \\tfrac12 x_2, \\ldots)$, half the right shift.\nThen $\\norm{T} = \\tfrac12$, so $I - T$ is invertible with $\\norm{(I - T)^{-1}} \\le 2$.\n\nCompleteness is needed.\nOn the space of finitely supported sequences with the $\\ell^2$ norm, $T$ as above has norm $\\tfrac12$.\nBut the solution of $(I - T)x = (1, 0, 0, \\ldots)$ is $x = (1, \\tfrac12, \\tfrac14, \\ldots)$, which is not finitely supported, so $I - T$ is not surjective there.\n\nA consequence: if $A$ is invertible and $\\norm{B - A} \u003c 1 / \\norm{A^{-1}}$, then $B$ is invertible, so the invertible operators form an open set.\nThe series also converges, with the same sum, whenever $\\norm{T^k} \u003c 1$ for some $k \\ge 1$.\n\n## Explanations\n\n### L1 human\n\nFor numbers, $1/(1 - t) = 1 + t + t^2 + \\cdots$ when $\\abs{t} \u003c 1$.\nThe Neumann series says the same formula works for operators, with the absolute value replaced by the operator norm.\nSo an operator that is a small perturbation of the identity can always be undone, and the inverse is computed by a convergent series.\n\n### L2 human\n\nThe Neumann series is the main tool to prove that invertibility is stable under small perturbations.\nIt shows that the set of invertible operators is open and that inversion is continuous, indeed analytic, on it.\nIn spectral theory it shows that the spectrum of $T$ lies in the closed disc of radius $\\norm{T}$, and that the resolvent is analytic.\nFor integral equations $x - Kx = y$ with a small kernel $K$, the series gives the solution by iteration.\nThe series is named after Carl Neumann, who used it in potential theory.\n",
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