I want to find the spectrum of operator J in $l^2$.
$(J u)_{n}= \begin{cases}b_{1} u_{1}+a_{1} u_{2} & n=1 \\ a_{n-1} u_{n-1}+b_{n} u_{n}+a_{n} u_{n+1} & n \geq 2\end{cases}$.
If $a_n=a,b_n=b$, then $\sigma(J)=[b-2a,b+2a]$,it has no point spectrum .
If $b_1=c,b_n=b$ when $n >1$ and $a_n=a$,where a,b,c is real number, Can it still has no point spectrum? if it has point spectrum,I guess it has only one eigenvalue (depend on a,b,c), but I cannot figure it out.
where is the transition line (about a,b,c) between $\sigma(J)$has no point spectrum and has point spectrum?
My attempts was similar to the question Jacobi operator's spectrum in $l_2$
First of all I tried to find point spectrum $\sigma_{p}(J)=\{\lambda \in \mathbb{C}: \operatorname{ker}(J-\lambda I) \neq\{0\}\}$, where $I$ is an identity operator. Let $x=\left(x_{1}, x_{2}, \ldots\right) \in l_{2} .$ We obtain equations of the form $$ J x=\lambda x \Leftrightarrow\left\{\begin{array}{l} x_{2}=\frac{(\lambda-b) x_{1}}{a} \\ x_{3}=\frac{(\lambda-b) x_{2}}{a}-x_{1} \\ x_{4}=\frac{(\lambda-b) x_{3}}{a}-x_{2} \\ \cdots \\ x_{n}=\frac{(\lambda-b) x_{n}-1}{a}-x_{n-2} \\ \cdots \end{array}\right. $$ Also we can obtain the equations for $x_{n}$ in the form $$ x_{n}=p_{n}\left(\frac{\lambda-b}{a}\right) x_{1} $$ where $p_{n}(x)$ is a polynomial of degree $n-1$. But the form of polynomials is remains unclear. Also this sequnce $x$ should belongs to $l_{2}$, that is $$ \sum_{n \geq 1}\left|x_{n}\right|^{2} \leq \infty $$