Positive definiteness of infinite tridiagonal matrices I am interested in the following problem: I have an infinite symmetric tridiagonal matrix 
$$
A=
\begin{bmatrix}
a_1 & b_1 &      &      &  \\
b_1 & a_2 & b_2 &      & \\
    & b_2& a_3  & b_3 & \\ 
    &    & \ddots & \ddots & \ddots & \\
\end{bmatrix}
$$
where $a_j, b_j>0$, and I need to determine whether $A$ is positive definite, meaning that the corresponding quadratic form is bounded below:
$$
Q_A(\beta_1, \beta_2\ldots \beta_n\ldots)\stackrel{\mathrm{def}}{=}\sum_{j=1}^\infty a_j \beta_j^2 + 2b_j\beta_{j}\beta_{j+1} \ge c\sum_{j=1}^\infty \beta_j^2.$$
Here $c>0$. (If $c=0$, we say that $A$ is positive semidefinite).

Question Are there infinite-dimensional versions of the familiar criterions of linear algebra, such as the Sylvester's criterion or the diagonal dominance sufficient condition? 

Any result or reference is gladly welcome.
 A: Presumably you mean $2 b_j$, not $b_j/2$.  
The appropriate context for this is
linear operators on $\ell^2$.  I'll just consider the case where the $a_j$ and $b_j$ are bounded, which makes $A_\infty$ correspond to a bounded self-adjoint 
linear operator $A$ on $\ell^2$.
If $P_n$ is the orthogonal projection on 
the span of the first $n$ standard unit vectors, Sylvester's criterion
essentially says that $P_n A P_n$ should be positive definite for all $n$.
This does imply that $A$ is positive semidefinite, but it is not necessarily
(strictly) positive definite: it may have a null space containing a 
vector with infinitely many nonzero entries.  For example, try
$$ A_\infty = \left[ \matrix{ 1/2 & 1 &  & & \cr
                              1   & 5/2 & 1 & &\cr
                                  & 1   & 5/2 & 1 & \cr
                                  &   & \ldots & \ldots & \ldots \cr}\right],
\ v = \left[\matrix{ 1\cr -1/2\cr 1/4\cr \ldots}\right] $$
A: Here's an easy answer based on the diagonal dominance criterion.
If the entries of $A$ satisfy the inequalities 
\begin{equation}
  \begin{cases}
    a_j\ge b_{j-1}+b_j + c, & j\ge 2 \\
    a_1\ge b_1 + c 
\end{cases}
\end{equation}
where $c\ge 0$, then the quadratic form $Q_A$ satisfies 
\begin{equation}
Q_A(\beta_1,\beta_2,\beta_3\ldots) \ge c\sum_{j=1}^\infty \beta_j^2.
\end{equation}
Proof:
\begin{equation}
\begin{split}
\sum_{j=1}^\infty a_j\beta_j^2+2b_j\beta_j\beta_{j+1} &= a_1\beta_1^2+2b_1\beta_1\beta_2 + \sum_{j=2}^\infty  a_j\beta_j^2+2b_j\beta_j\beta_{j+1}\\
&\ge b_1\beta_1^2+2b_1\beta_1\beta_2+b_1\beta_2^2+\sum_{j=2}^\infty b_j\beta_{j+1}^2+b_j\beta_j^2+2b_j\beta_j\beta_{j+1} + c\sum_{j=1}^\infty\beta_j^2 \\
&=\sum_{j=1}^\infty b_j\left(\beta_j+\beta_{j+1}\right)^2+c\sum_{j=1}^\infty \beta_j^2.
\end{split}
\end{equation}
