Why does $(I-T)S_k-I$ equal $I-T^{k+1}-I$? Consider a Banach space, $X$, and $T\in\mathcal B(X)$ where $\|T\|\lt1$. Take as well that $S=\sum_{n=0}^\infty T^n$ so that it's $k$th partial sum is given by $S_k=\sum_{n=0}^k T^n$. Suppose as well that $\{S_k\}$ converges to $S$ in $B(X)$.
Why is it that,
$$(I-T)S_k-I=I-T^{k+1}-I$$
and not that,
$$(I-T)S_k-I=S_k-T^{k+1}-I\,?$$
(This is part of the proof of the Neumann series theorem, which says that under the above conditions, $I-T$ is an invertible operator and it's inverse is given by $\sum_{n=0}^\infty T^n$).

I'm going to write in lowercase. 
\begin{align}
(1-t)s_k  - 1 
&= s_k - ts_k - 1 \\
&= 1 + t + \ldots + t^k- t(1 + t + \ldots + t^k) - 1 \\
&= 1 + t + \ldots + t^k- (t + t^2 + \ldots + t^{k+1}) - 1 \\
&= 1 + t + \ldots + t^k- t - t^2 + \ldots - t^{k+1} - 1 & \text{cancel common terms}\\
&= 1  - t^{k+1} - 1.
\end{align}
