<s>[INST] Maximum eigenvalue of a covariance matrix of Brownian motion $$ A := \begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{3} & \cdots & \frac{1}{n}\\
\frac{1}{2} & \frac{1}{2} & \frac{1}{3} & \cdots & \frac{1}{n}\\
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} & \cdots & \frac{1}{n}\\
\vdots & \vdots & \vdots & \ddots & \frac{1}{n}\\
\frac{1}{n} & \frac{1}{n} & \frac{1}{n} & \frac{1}{n} & \frac{1}{n}
\end{pmatrix}$$
How to prove that all the eigenvalues of $A$ are less than $3 + 2 \sqrt{2}$?
This question is similar to this one.
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 $1, 1/2, 1/3, \ldots, 1/n$
 [/INST] Inspired by @Mateusz Wasilewski I find another method.
\begin{eqnarray*}
\langle x,Ax\rangle & = & \langle Lx,Lx\rangle\\
 & = & \sum_{i=1}^{n}u_{i}^{2}
\end{eqnarray*}
where $u_{i}=\sum_{j=i}^{n}\frac{1}{j}x_{j}$.
\begin{eqnarray*}
\sum_{i=1}^{n}u_{i}^{2} & = & \sum_{i=1}^{n}(\sum_{k=i}^{n}b_{k})^{2}\quad(\text{where} \ b_{k}=\frac{1}{k}x_{k})\\
 & = & \sum_{i=1}^{n}(\sum_{k=i}^{n}b_{k}^{2}+2\sum_{k>j\geq i}b_{k}b_{j})\\
 & = & \sum_{k=1}^{n}\sum_{i=1}^{k}b_{k}^{2}+2\sum_{j=1}^{n-1}\sum_{k=j+1}^{n}\sum_{i=1}^{j}b_{k}b_{j}\\
 & = & \sum_{k=1}^{n}\frac{x_{k}^{2}}{k}+2\sum_{j=1}^{n-1}\sum_{k=j+1}^{n}b_{k}x_{j}\\
 & = & \sum_{k=1}^{n}\frac{x_{k}^{2}}{k}+2\sum_{k=2}^{n}\sum_{j=1}^{k-1}b_{k}x_{j}\\
 & = & \sum_{k=1}^{n}\frac{x_{k}^{2}}{k}+2\sum_{k=2}^{n}b_{k}z_{k-1}\\
 & = & x_1^2 +\sum_{k=2}^{n}\frac{(z_{k}-z_{k-1})^{2}}{k}+2\sum_{k=2}^{n}\frac{(z_{k}-z_{k-1})}{k}z_{k-1}\\
 & = & x_{1}^{2}+\sum_{k=2}^{n}\frac{z_{k}^{2}-z_{k-1}^{2}}{k}\\
 & = & \sum_{k=1}^{n}\frac{z_{k}^{2}}{k}-\sum_{k=1}^{n-1}\frac{z_{k}^{2}}{k+1}\\
 & = & \sum_{k=1}^{n-1}z_{k}^{2}(\frac{1}{k}-\frac{1}{k+1})+\frac{z_{n}^{2}}{n}
\end{eqnarray*}
</s>