$$ A := \begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{3} & \cdots & \frac{1}{n}\\
\frac{1}{2} & \frac{1}{2} & \frac{1}{3} & \cdots & \frac{1}{n}\\
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} & \cdots & \frac{1}{n}\\
\vdots & \vdots & \vdots & \ddots & \frac{1}{n}\\
\frac{1}{n} & \frac{1}{n} & \frac{1}{n} & \frac{1}{n} & \frac{1}{n}
\end{pmatrix}$$


How to prove that all the eigenvalues of $A$ are less than $3 + 2 \sqrt{2}$?


This question is similar to [this one](https://mathoverflow.net/q/45982).


I have tried the Cholesky decomposition $A = L^{T} L$, where


$$L^{T} = \left(\begin{array}{ccccc}
1 & 0 & 0 & \cdots & 0\\
\frac{1}{2} & \frac{1}{2} & 0 & \cdots & 0\\
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} & \cdots & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
\frac{1}{n} & \frac{1}{n} & \frac{1}{n} & \frac{1}{n} & \frac{1}{n}
\end{array}\right)$$


then


$$(L^{T})^{-1}=\left(\begin{array}{ccccc}
1 & & & \cdots\\
-1 & 2 & & \cdots\\
 & -2 & 3 & \cdots\\
\vdots & \vdots & \vdots & \ddots\\
 & & & -(n-1) & n
\end{array}\right)$$


$$A^{-1}=L^{-1}(L^{T})^{-1}$$


How to prove the eigenvalues of $A^{-1}$


$$\lambda\_{i}\geq\frac{1}{3+2\sqrt{2}}$$


Further, I find that $A$ is the covariance matrix of Brownian motion at time $1, 1/2, 1/3, \ldots, 1/n$



Maximum eigenvalue of a covariance matrix of Brownian motion
In this answer I show that the largest eigenvalue is bounded by $5< 3 + 2\sqrt{2}$. I will first use the interpretation of this matrix as the covariance matrix of the Brownian motion at times $(\frac{1}{n},\dots, 1)$ (I reversed the order so that the sequence of times is increasing, which is more natural for me).


We have $A\_{ij} = \mathbb{E} (B\_{t\_{i}} B\_{t\_j})$. The largest eigenvalue will be the supremum over the unit ball of the expression $\langle x, A x\rangle$, which is equal to $\sum\_{i,j} A\_{ij} x\_{i} x\_{j}$. This is equal to $\mathbb{E} (\sum\_{i=1}^{n} x\_{i} B\_{t\_{i}})^2$. In order to exploit the independence of increments of the Brownian motion, we rewrite the sum $\sum\_{i=1}^{n} x\_i B\_{t\_{i}}$ as $\sum\_{i=1}^{n} y\_{i} (B\_{t\_{i}} - B\_{t\_{i-1}})$, where $y\_{i}:= \sum\_{k=i}^{n} x\_{k}$ and $t\_0:=0$. Thus we have


$
\mathbb{E} (\sum\_{i=1}^{n} x\_{i} B\_{t\_{i}})^2 = \sum\_{i=1}^{n} y\_{i}^2 (t\_{i}-t\_{i-1}).
$


The case $i=1$ is somewhat special and its contribution is $\frac{y\_1^2}{n} \leqslant \sum\_{k=1}^{n} x\_{k}^2 = 1$. For the other ones we have $t\_{i} - t\_{i-1} = \frac{1}{(n-i+1)(n-i+2)}\leqslant \frac{1}{(n-i+1)^2}$. At this point, to get a nicer expression, I will reverse the order again by defining $z\_{i}:= y\_{n-i+1}$. So we want to estimate the expression


$
\sum\_{i=1}^{n} \left(\frac{z\_i}{i}\right)^2.
$


We can now use use [Hardy's inequality](https://en.wikipedia.org/wiki/Hardy%27s_inequality) to bound it by $4 \sum\_{i=1}^{n} x\_{i}^2 =4$. So in total we get 5 as an upper bound, if I haven't made any mistakes.


