Another mystery from Ulrich Krengel's textbook - Ergodic Theorems (first mystery). This time it's from page 190, in the proof of theorem 2.7.
He takes $P=T^2$, where $T$ is a self-adjoint positive contraction in $L^2$, and argues that it's easy to check that:
$$P^n f - A_{n+1}(P) f = (n+1)^{-1}\sum_{k=1}^n k(P^k - P^{k-1})f$$
where $A_{n+1}(P) f = (n+1)^{-1} \sum_{k=0}^n P^k f$.
I only get that:
$$\frac{n}{n+1}P^n f -(n+1)^{-1} f -\sum_{k=1}^{n-1}(n+1)^{-1}P^k f$$
Can someone show me how did he get this sum above?
Starting from where you get and calling $x_k=P^kf$, one must show that $$nx_n-x_0 -\sum_{k=1}^{n-1}x_k=\sum_{k=1}^n k(x_k - x_{k-1}).$$ This is purely algebraic. For example, the RHS is $$\sum_{k=1}^n kx_k -\sum_{k=1}^n k x_{k-1}=\sum_{k=1}^n kx_k -\sum_{k=0}^{n-1}(k+1) x_{k}=nx_n+\sum_{k=1}^{n-1}kx_k -x_0-\sum_{k=1}^{n-1}(k+1) x_{k},$$ that is, $$nx_n-x_0-\sum_{k=1}^{n-1}x_{k}=(n+1)x_n-\sum_{k=0}^{n}x_{k},$$ as desired.