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 $\frac{y_1^2}{n} \leqslant \sum_{k=1}^{n} y_{k}^2 = 1$. 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 to estimate it above 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.

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 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.