$I+T$ is not bounded below Let $\mathcal{H}$ a separable Hilbert space. I want to show that the operator $I+S \in L(\mathcal{H})$, with $S\in L(\mathcal{H})$ the left shift, is not bounded below; i.e, there exists a sequence $\{x_n\}_{n\in \mathbb{N}}\subset \mathcal{H}$ such that $||x_n||=1$ and $T(x_n)\underset{n \longrightarrow \infty}{\longrightarrow} 0$.
My aproach was:
I couldn't find the suitable sequence, then i tried the following...
$\sigma(S)=\mathbb{D} \implies S+I \notin \mathcal{G}_l(\mathcal{H})$. That's implies $I+S$ isn't bounded below or $I+S^*$ isn't bounded below. I tried to find a lower bound for $I+S^*$, but i couldn't.
Thanks for read.

For each $n\in\Bbb N$ define $x_n=(x_{n,k})_k\in\mathcal H=\ell^2(\Bbb N)$ by
$$x_{n,k}:=\begin{cases}\frac{(-1)^k}{\sqrt n}&\text{ if }1\le k\le n\\0&\text{ if }k>n.\end{cases}$$
Then, $\|x_n\|=1,$ and $T(x_n)\to0$ since
$$T(x_n)_k:=\begin{cases}\frac{(-1)^n}{\sqrt n}&\text{ if }k=n\\0&\text{ else.}\end{cases}$$
