Is there a sharper upper bound of the spectral norm of Hilbert matrix than $3+2\sqrt{2}$? 
$A_n$ is a real symmetric $n \times n$ matrix defined by
$$ A_n =  \begin{bmatrix} 1 & \frac{1}{2} & \frac{1}{3} & \cdots &
 \frac{1}{n} \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{3} & \cdots &
 \frac{1}{n} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & \cdots &
 \frac{1}{n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\
 \frac{1}{n} & \frac{1}{n} & \frac{1}{n} & \cdots & \frac{1}{n} \\
 \end{bmatrix} $$
Find an upper bound of the eigenvalues of $A_n$ as tight as possible.


Let $A_n = U+U^T-D,$ where $U$ is the upper triangular part of $A_n$ and $D$ is the diagonal of $A_n$.
Suppose $\lambda$ is the largest eigenvalue of $A_n = U^T U$. 
For any $X \in \mathbb{R}^n$, 
$$X^T A_n X = X^T U^T U X = \left\Vert UX \right\Vert^2 \leq \lambda \left\Vert X \right\Vert^2 \implies \left\Vert UX \right\Vert \leq \sqrt{\lambda} \left\Vert X \right\Vert.$$
Since $U^T U$ is similar to $U U^T$, $\lambda$ is also the largest eigenvalue of $ U U^T $. 
Hence we also have
$$ \left\Vert U^T X \right\Vert \leq \sqrt{\lambda} \left\Vert X \right\Vert.$$
Similarly, $$ \left\Vert D X \right\Vert \leq \left\Vert X \right\Vert.$$
Thus 
$$ \left\Vert A_n X \right\Vert = \left\Vert U X + U^T X - D X \right\Vert \leq \left\Vert U X \right\Vert + \left\Vert  U^T X \right\Vert + \left\Vert  D X \right\Vert \leq \left( 2 \sqrt{\lambda} + 1 \right) \left\Vert X \right\Vert.$$
Take $X$ to be the eigenvector of $A_n$ corresponding to $\lambda$, then
$$ \left\Vert A_n X \right\Vert = \lambda \left\Vert X \right\Vert \leq \left( 2 \sqrt{\lambda} + 1 \right) \left\Vert X \right\Vert,$$
we have $\sqrt{\lambda} \leq 1+\sqrt{2} \implies \lambda \leq 3+2\sqrt{2}.$

Is there any tighter upper bound?
 A: The matrix $A_n$ is semi-famous because its inverse is a tridiagonal matrix: $$A_n^{-1}=2B_n=2\begin{pmatrix}
1 & -1 & 0 & \cdots & 0 \\
 -1 & 4 & -3 & 0 & \\
 0 & \ddots & \ddots & \ddots & 0 \\
 0 & 0 & -t_{n-2} & (n-1)^2 & -t_{n-1} \\
 0 & 0 & 0 & -t_{n-1} & n^2/2\end{pmatrix}$$ where $t_n=n(n+1)/2$ are the triangular numbers. It seems easier to find the smallest eigenvalue of $B_n$ than the largest eigenvalues of $A_n$. If $v=(u_1(x),u_2(x),\ldots,u_n(x))$ is an eigenvector of $B_n$ with eigenvalue $x$, then the components satisfy the recurrence relation $$u_{k+1}(x)=\frac{k^2-x}{t_k}u_k(x)-\frac{t_{k-1}}{t_k}u_{k-1}(x),\quad u_1(x)=1,u_2(x)=1-x$$ $$\left(\frac{n^2}{2}-x\right)u_n(x)=t_{n-1}u_{n-1}(x)\tag{1}$$ the last being the characteristic polynomial equation of $B_n$. To find the smallest root of this polynomial one might use more sophisticated techniques, but for brevity let's keep only the constant and $x$ term. The proofs are omitted but it is not hard to see that $$u_n(x)=1-2(H_n-1)x+\cdots$$ where $H_n$ are the harmonic numbers. Substituting these into $(1)$ and neglecting higher order terms gives $x\approx \frac{n}{2 \left(n H_{n-1}+1\right)}$, so the largest eigenvalue of $A_n$ is less than $$\lambda\le H_n$$ A comparison of the actual largest eigenvalue and $H_n$ is below: $$\begin{matrix}n&1&2&3&4&5&\ldots&20\\
\lambda_n&1&1.31&1.49&1.62&1.71&&2.22\\
H_n&1&1.5&1.83&2.08&2.28&&3.66\\
\end{matrix}$$
A: The matrix $A_{n}$ is known due to its relation to the Hilbert matrix and also the finite section Hardy inequality. Its operator norm, which coincides with its largest eigenvalue, is the best constant in the Hardy inequality.
A tighter and the best bound independent of $n$ reads
$$
\|A_{n}\|\leq4
$$
since $4$ equals the norm of its semi-infinite variant (sometimes referred to as the Hilbert $L$-matrix) regarded as an operator on $\ell^{2}(\mathbb{N})$.
For $n\to\infty$, an asymptotic expansion of $\|A_{n}\|$ can be deduced in terms of negative powers of $\log n$ up to an arbitrary order. For example, first three terms are as follows:
$$
 \|A_{n}\|=4-\frac{16\pi^{2}}{\log^{2}n}+\frac{32\pi^{2}(\gamma+6\log 2)}{\log^{3}n}+O\left(\frac{1}{\log^{4}n}\right),
$$
where $\gamma$ is the Euler--Mascheroni constant. Proofs and references can be found in this paper.
I don't think that an $n$-dependent improvement of the upper bound for $\|A_{n}\|$ is known but one could guess from the asymptotic formula that an inequality of the form
$$
 \|A_{n}\|\leq 4-\frac{c}{\log^{2} n}
$$
could be true for some $c>0$ and $n\geq2$ (or maybe few more first values of $n$ excluded).
