Question: I saw this definition in one of the computer science book but I am unable to recall the theorem name. Can someone please provide the reference?

$$\lim\_{n \to \infty} \frac{f(n)}{g(n)} = c > 0$$ means there is some $n\_{0}$ beyond which the ratio is always between $\frac{1}{2}c$ and $2c$.

Answer: 
Notice that $\|Ax\| \le \|x\|$. So $\sigma(A)$ is contained in the closed unit disk $D^{c}$. To find the resolvent $(A-\lambda I)^{-1}$, try to solve $(A-\lambda I)x = y$ for $x$, and see if it can be done. First, see if there are any eigenfunctions $Ax=\lambda x$. If there were such an $x$, then $|\lambda| \le 1$ and $x=(x\_1,x\_2,x\_3,\cdots)$ would satisfy
$$
 (x\_2-\lambda x\_1,x\_3-\lambda x\_2,x\_4-\lambda x\_3,\cdots)=0,\\
 x\_2=\lambda x\_1,\; x\_3=\lambda x\_2,\;x\_4=\lambda x\_3.
$$
Assuming $x\_1=1$, there is a solution $(1,\lambda,\lambda^{2},\lambda^{3},\cdots)$. This solution is valid only for $|\lambda| < 1$ because, otherwise, $x \notin l^{2}$. In other words, $\sigma\_{P}(A)=\{ \lambda \in \mathbb{C} : |\lambda| < 1\}$ is the open unit disk. And $\sigma(A)=\{ \lambda \in\mathbb{C} : |\lambda|\le 1\}$ because $\sigma(A)$ is closed, contains the open unit disk, and is contained in the closed unit disk.

The adjoint of $A$ is the shift $B(x\_1,x\_2,x\_3,\cdots)=(0,x\_1,x\_2,x\_3,\cdots)$. Notice that $\|Bx\|=\|x\|$ for all $x \in l^{2}$. $B$ has no point spectrum because $Bx=\lambda x$ implies $(x\_1,x\_2-\lambda x\_1,x\_3-\lambda x\_2,\dots)=0$, which forces $x\_1=0, x\_2=\lambda x\_1=0,etc.$. So the range of $(A-\lambda I)$ is dense for all $\lambda$, which puts the unit circle $\{ \lambda \in\mathbb{C} : |\lambda|=1\}$ in the continuous spectrum of $A$. $A$ has no residual spectrum.
Since $\|A\| = 1$, using the Neumann series we can indeed invert $A - \lambda I$ when $|\lambda|>1$. For $|\lambda|<1$, the sequence $x\_n = \lambda^n$ lies in the kernel of $A-\lambda I$ and it is not invertible.

Then the conclusion follows by recalling that the spectrum of a bounded operator is always closed.