Today I was asked by a friend to have a look over a linear algebra problem, which is quite easy at a first glance:
For $n\in\mathbb N$, put $$A_n=\left[\begin{matrix}1 &\frac{1}{2} &\frac{1}{3} &\cdots &\frac{1}{n}\\ \frac{1}{2} &\frac{1}{2} &\frac{1}{3} &\cdots &\frac{1}{n}\\ \frac{1}{3} &\frac{1}{3} &\frac{1}{3} &\cdots &\frac{1}{n}\\ \vdots& & &\ddots & \vdots\\ \frac{1}{n}& \frac{1}{n}& \frac{1}{n}& \cdots &\frac{1}{n}\end{matrix}\right].$$ Then the eigenvalues of $A_n$ are all positive and no more than $3+2\sqrt2$.
To the positivity of eigenvalues, we may consider the matrix $P=\left[\begin{smallmatrix} 1 \\ -1 & 1\\ & \ddots& \ddots \\& & & 1\\& & & -1 &1\end{smallmatrix}\right]$ and it follows that $P^\top AP=\left[\begin{smallmatrix} 1/2 \\ & 1/6& &\ast \\ & &\ddots\\ & & & 1/(n-1)n\\ & O& & & 1/n\end{smallmatrix}\right]$.
Now the problem is, the wierd bound $3+2\sqrt 2$ cannot be seen. Diagonalizing this real symmetric matrix does not seem to work because it usually needs the eigenvalues beforehand. I thought about Gerschgorin disk but unfortunately it does no help here. So I would like to ask for some ideas about this problem, and any help is appreciated.
Actually all the eigenvalues of $A_n$ are less than $4$ [see Section V of Man-Duen Choi's paper Tricks or Treats with the Hilbert Matrix, American Mathematical Monthly 90 (1983), 301-312]. Because $3 + 2 \sqrt{2} \approx 5.8$, the claimed result is correct but is not the best possible upper bound of the eigenvalues of $A_n$.