On Fredholm operator Consider operator $T: l^2(\mathbb{N})\to l^2(\mathbb{N})$ given by $T(x_1,x_2,\cdots)=(\lambda_1x_1,\lambda_2x_2,\cdots)$, where $\{\lambda_n\}_{n\in \mathbb{N}}$ is nonzero bounded complex numbers. I was asked to show if $\liminf_{n \to \infty} |\lambda_n|>0$, then $T$ can be written into the sum of a compact operator and an invertible operator, thus Fredholm. 
[Some observations]
It is quite clear that if $\lim_{n \to \infty} \lambda_n=0$, then $T$ is a compact operator, and since we can rotate $\lambda_n$ on complex plane by its argument counterclockwise, we may assume $\lambda_n>0$. Now the problem is reduced to construct two operators from given $\{\lambda_n\}_{n\in \mathbb{N}}$, one invertible and the other compact. I tried to construct a compact operator from one convergent subsequence of $\{\lambda_n\}_{n\in \mathbb{N}}$ but then rapped to modify another into an invertible operator. 
 A: When $c := \liminf\limits_{n\to\infty} \lvert \lambda_n\rvert > 0$, then $T$ is itself already invertible, as the $\lambda_n$ are assumed to be nonzero. Because there is an $n_0 \in \mathbb{N}$ with $\lvert \lambda_n\rvert > c/2$ for all $n \geqslant n_0$, and $\varepsilon := \min \{ \lvert \lambda_k\rvert : k \leqslant n_0\} > 0$ by the assumption $\lambda_n \neq 0$. Then $T^{-1}$ is given by the pointwise multiplication with the bounded (by $\max \{2/c,\, 1/\varepsilon\}$) sequence $(\lambda_n^{-1})_{n\in\mathbb{N}}$.
If the assumption $\lambda_n \neq 0$ is dropped, then the addition of a nonzero compact operator is in general necessary, let $A = \{ n \in \mathbb{N} : \lambda_n = 0\}$ and
$$P_A(x) = (y_k),\quad y_k = \begin{cases}x_k &, k\in A\\0 &, k \notin A\end{cases}$$
the projection onto the subspace determined by $A$. Then $T = (T-P_A) + P_A$, where $T-P_A$ is invertible and $P_A$ is compact since it has finite dimensional range.
More generally, for $\mu \in \ell^\infty(\mathbb{N})$, the multiplication operator
$$M_\mu \colon \ell^2(\mathbb{N}) \to \ell^2(\mathbb{N});\quad M_\mu(x) = (\mu_k\cdot x_k)$$
has the properties that $M_\mu - \lambda I$ is invertible if and only if $\lambda \notin \overline{\{\mu_k : k \in \mathbb{N}\}}$, and $M_\mu - \lambda I$ can be written as the sum of an invertible operator and a compact operator if and only if $\lambda$ is not a limit point of the sequence $\mu$.
The construction of an invertible operator $S$ and a compact $K$ such that $M_\mu -\lambda I = S + K$ when $\lambda$ is not a limit point of $\mu$ is like above.
Conversely, suppose $\lambda$ is a limit point of $\mu$, and there were a decomposition
$$M_\mu - \lambda I = S + K$$
with invertible $S$ and compact $K$. Since $\lambda$ is a limit point of $\mu$, there is a subsequence $\mu_{n_k} \to \lambda$, and hence
$$(M_\mu - \lambda I)(e_{n_k}) \to 0,$$
where $(e_n)$ is the obvious ONB of $\ell^2$. Since $K$ is compact, there is a subsequence with
$$K(e_{n_{k_m}}) \to \xi.$$
Then
$$\eta_m = S(e_{n_{k_m}}) = (M_\mu - \lambda I - K)(e_{n_{k_m}}) \to -\xi,$$
but
$$e_{n_{k_m}} = S^{-1}(\eta_m) \not\to S^{-1}(-\xi),$$
contradicting the continuity of $S^{-1}$.
