Prove that $A^n$ converges to the null operator $A \in \mathcal B(H)$, $||A||<1$

Question

I have an operator $A \in \mathcal B(H)$ where $H$ in a Hilbert space, $||A||<1$. I have to show that:

1. $A^n$ converges to the null operator
2. $S_n=\sum_{k=0}^n A^k$ converges
3. $S_n \to (1-A)^{-1}, n \to\infty$

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I don't know how to approach the first request. I've tried, without certainty: $$||A^n||=\sup_x \frac{||A^nx||}{||x||}=\sup_{||x||=1} {||A^nx||} \le \sup_x|||A||^n| \to 0$$

I can answer to questions 2. and 3. with the proof of the following theorem:

If $X$ is a Banach space, $A \in \mathcal B(X)$, $|A|<1$, then $(1-A)^{-1}$ exists in $\mathcal B(X)$ and it's given by the Neumann series: $$(1-A)^{-1}=\sum_{k=0}^\infty A^k$$

Answer 1

$\|A(Ax)\|\leq\|A\|\|Ax\|\leq\|A\|\|A\|\|x\|=\|A\|^{2}\|x\|$, so $\|A^{2}\|\leq\|A\|^{2}$, and the rest $n$ is just by induction.