Hi all, 


do you know how to compute (as a function of n) the largest eigenvalue of this matrix (or at least to bound it)?


$$
\left(\begin{array}{cccccc}
0 & 1 & & & & \cr
1 & 0 & \sqrt 2 & & & \cr
 & \sqrt 2 & 0 & & & \cr
 & & & \ddots & & & \cr
 & & & & 0 & \sqrt n & \cr
 & & & & \sqrt n & 0 &
\end{array}\right)
$$


Thanks!



Bounding largest eigenvalue
If you denote $A\_n$ your tri-diagonal matrix of order $n$, and $H\_n(x):= \det(x+A\_n)$, the sequence $H\_n$ satisfies the two-term linear recurrence $H\_{n+1}=xH\_n - nH\_{n-1}$ with initial conditions $H\_0=1$ and $H\_1=x$. Thus, they are the [Hermite polynomials](http://en.wikipedia.org/wiki/Hermite_polynomials) (here in the "probabilist's version"), and their zeros are the eigenvalues of $-A\_n$ (on which you can find everything in the literature).


