Show that $\frac{1}{n}\sum_{j=1}^n T^jx$ converges. $T: \ell^2\to \ell^2$ with $T(x_n) = (a_nx_n)_{n\geq 1}$, $\sup \lvert a_n \rvert \leq 1$

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I'm working in "Operator Theoretic aspects of Ergodic theory" by Eisner, Farkas Haase and Nagel. The question is as follows

Consider the Hilbert space $H := \ell^2(\mathbb{N})$ and a sequence $(a_n)_{n\in\mathbb{N}}$ in $\mathbb{C}$ with $\sup_{n\in\mathbb{N}}\lvert a_n \rvert\leq 1$. Define $T: H\to H$ by $T(x_n)_{n\in\mathbb{N}} = (a_nx_n)_{n\in\mathbb{N}}$. Prove that for every $x = (x_n)_{n\in\mathbb{N}}\in H$ \begin{align} \frac{1}{n}\sum_{j=1}^n T^jx \quad \text{converges as $n\to \infty$} \label{ch1:eq:avg_sum} \end{align} and determine the limit.

My attempted solution is as follows:

It is easy to see that $T$ maps in fact into $\ell^2$.

Now, we also have to find the limit. First, observe that

$$ T^jx = \underbrace{(T\circ T\circ \ldots \circ T)}_{j\text{ times}}x = (a_1^jx_1, a_2^jx_2,\ldots). $$ Then, the sum in $$ \frac{1}{n} \sum_{j=1}^n T^jx = (\frac{a_1 + a_1^2 + \ldots + a_1^n}{n}x_1, \frac{a_2 + a_2^2 + \ldots + a_2^n}{n}x_2, \ldots ) $$ We need to find an element $y\in H$ such that $\lVert\frac{1}{n} \sum_{j=1}^n T^jx - y\rVert_{\lt}\to 0$ as $n\to \infty$. This is the point where I am stuck. The most ``natural'' choice of limit would be $y = x$. However, then we need to show the following.
Let $y = x$, then we have \begin{align} \lVert\frac{1}{n} \sum_{j=1}^n T^jx - x\rVert_{\ell^2}^2 &= \sum_{j=1}^n \lvert\frac{a_j + a_j^2 + \ldots + a_j^n}{n} - 1\rvert^2 \lvert x_j\rvert^2 + \sum_{k = n+ 1}^\infty \lvert x_k\rvert^2 \end{align}

We call $$ S_{n,j} = \frac{a_j + a_j^2+\ldots + a_j^n}{n} = \frac{1 + a_j + a_j^2+\ldots + a_j^n}{n} -\frac{1}{n} = \frac{1}{n}\left(\sum_{k=0}^n a_j^k - 1\right). $$ The sum is a partial sum of a geometric series, which equals $$ \sum_{k=0}^n a_j^k = \frac{a_j^{n+ 1} - 1}{a_j - 1} \implies S_{n,j} = \frac{1}{n}(\frac{a_j^{n+ 1} - 1}{a_j - 1} - 1). $$

But from here I am stuck as I do not know how to show that x is the right limit. Which I don't think it is. Could someone give a hint towards the right direction? Thank you!