$H$ is a separable Hilbert space over $\mathbb C$ and $\{u_n\}$ is a maximal orthonormal set of H. $A \in B(H)$ and there exists $\lambda \in \mathbb C$ such that $$A(u_n) = \lambda u_n - u_{n+1}, n = 1,2,...$$ I need to evaluate $\sigma(A)$.
I can see that $(A-\mu I)(u_n) = (\lambda - \mu)u_n - u_{n+1}$. If $\mu = \lambda$ it seems that the $A - \lambda I$ is not invertible as $u_1$ does not have a pre-image. How about $\mu \neq \lambda$. I do not know how to discuss it.
Also, why do we need the condition that $H$ is separable? I do not know where I need such information.
Let's first see if it has any eigenvalues. Let $x = (x_1,x_2,\ldots)\in H$ (in terms of the basis $\{u_n\}$ given above) and suppose that it is an eigenvector, then
$$Ax = \mu x \Longrightarrow (\lambda x_1 - x_2,\lambda x_2-x_3,\ldots) = \mu (x_1,x_2,\ldots).$$
Then $\lambda x_i - x_{i+1} = \mu x_i$, i.e. $x_{i+1} = (\lambda - \mu)x_i$. Inductively, this gives us
$$x = x_1(1,\lambda-\mu,(\lambda-\mu)^2,\ldots).$$
The only way for this to be in $H$ is if
$$\sum_{n=0}^{\infty} (\lambda-\mu)^n < \infty.$$
This is a geometric series and the only way for it to be finite is if $|\lambda-\mu| < 1$. This is an open disk centered at $\lambda$ of radius one. Recall that the spectrum of a bounded operator is compact, so it is closed. This should give you an idea of how to proceed.