Point on the proof that the inverse operator of $I-T$ is given by $(I-T)^{-1}=\sum_{k=0}^\infty T^k$

Question

Let $X$ be a Banach space and let $T\in B(X)$ be such that $\|T\|\lt1$. Suppose then we have the operator $I-T$ and we want to show that its inverse operator $(I-T)^{-1}$ is given by the following expression,

$$(I-T)^{-1}=\sum_{k=0}^\infty T^k$$

I have the following as proof of the above in my notes, but am unsure of a particular point.

Letting, $S=\sum_{k=0}^\infty T^k$ and $S_n=\sum_{k=0}^n T^k\,,n=1,2,...$, then $S_n\to S$ as $n\to\infty$.

So we consider,

$\|(I-T)S_n-I\|=\|I-T^{n+1}-I\|=\|-T^{n+1}\|\le\|T\|^{n+1}\to0\,,n\to\infty$

Why is it in the above that $I\cdot S_n=I$ and not $I\cdot S_n=S_n$?

Answer 1

It's not $I \cdot S_n=I$, it's a telescopic series: \begin{align} (I-T)S_n & = S_n - TS_n = \\ &= (T^0 + T^1 + \dots + T^n) - T(T^0 + T^1 + \dots + T^n)= \\ & = T^0 + T^1 + \dots + T^n - T^1 - T^2 - \cdots T^{n+1}= \\ & =T^0 - T^{n+1}=I-T^{n+1}. \end{align}

Answer 2

I assume you mean the transition $\|(I-T)S_n-I\|=\|I-T^{n+1}-I\|$?

This is basically the geometric series:

$$(I-T)S_n = S_n - T S_n = (I+T+...+T^n)-(T+T^2+...+T^{n+1}) = I-T^{n+1}$$