<s>[INST] Is this operator bounded? Is it compact? I'm working in a Hilbert space environment, say $H=l_2(\mathbb{N})$, with the underlying field of scalars $\mathbb{R}$. We know that if a linear operator on $H$ is bounded (i.e. continuous) then it has an infinite matrix representation. I have a linear and symmetric operator $J:H \longrightarrow H$ with matrix representation
$$\begin{pmatrix}
-1&\frac{1}{2}&\frac{1}{3}&\frac{1}{4}& \cdots\\\frac{1}{2}&-\frac{1}{2^2}&\frac{1}{6}&\frac{1}{8}& \cdots \\ \frac{1}{3}&\frac{1}{6}&-\frac{1}{3^2}& \frac{1}{12}&\cdots\\\frac{1}{4}&\frac{1}{8}&\frac{1}{12}&-\frac{1}{4^2}& \cdots\\\vdots&\vdots&\vdots&\vdots&\ddots
\end{pmatrix}$$
I would like to show that $J$ is bounded. It would be nice if $J$ was also compact. 
My initial intuition was that $J$ must be bounded, right? So I went straight for compactness: the initial strategy was to show that $J$ can be written as a product $J = SD$ and use the fact that a composition of a bounded operator and a compact operator is compact. 
$$J = \begin{pmatrix}
-1&\frac{1}{2}&\frac{1}{3}&\frac{1}{4}& \cdots\\\frac{1}{2}&-\frac{1}{2^2}&\frac{1}{6}&\frac{1}{8}& \cdots \\ \frac{1}{3}&\frac{1}{6}&-\frac{1}{3^2}& \frac{1}{12}&\cdots\\\frac{1}{4}&\frac{1}{8}&\frac{1}{12}&-\frac{1}{4^2}& \cdots\\\vdots&\vdots&\vdots&\vdots&\ddots
\end{pmatrix} =SD= \begin{pmatrix}
1&-1&-1&-1& \cdots\\ -\frac{1}{2}&\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}& \cdots \\ -\frac{1}{3}&-\frac{1}{3}&\frac{1}{3}& -\frac{1}{3}&\cdots\\ -\frac{1}{4}&-\frac{1}{4}& -\frac{1}{4}&\frac{1}{4}& \cdots\\\vdots&\vdots&\vdots&\vdots&\ddots
\end{pmatrix} \begin{pmatrix}
-1&0&0&0& \cdots\\ 0& -\frac{1}{2}&0&0& \cdots \\ 0&0&-\frac{1}{3}& 0&\cdots\\ 0&0&0&-\frac{1}{4}& \cdots\\\vdots&\vdots&\vdots&\vdots&\ddots
\end{pmatrix}$$
However, the operator $S$ above is not bounded. 
Another strategy to show compactness is to show that $J$ can be approximated by a sequence of finite rank operators $\{J_k\}$ where $||J_k-J|| \longrightarrow 0$ where $|| \cdot||$ is the operator norm. I suspect that a sequence of projection type operators such as
$$J_1 = \begin{pmatrix}
-1&0&0&0& \cdots\\ 0& 0&0&0& \cdots \\ 0&0&0& 0&\cdots\\ 0&0&0&0& \cdots\\\vdots&\vdots&\vdots&\vdots&\ddots
\end{pmatrix}, J_2 = \begin{pmatrix}
-1&\frac{1}{2}&0&0& \cdots\\ \frac{1}{2}& \frac{1}{2^2}&0&0& \cdots \\ 0&0&0& 0&\cdots\\ 0&0&0&0& \cdots\\\vdots&\vdots&\vdots&\vdots&\ddots
\end{pmatrix},...$$
may work, though I'm having trouble getting an upper bound on $||J_k-J||$. 
Thanks!
 [/INST] I'll freely identify operators with their matrices.
Let $v=(1,1/2,1/3, \dots).$ Let $A$ be the matrix whose $n$th column is $v/n.$ Let $B$ be the maxtrix whose $n$th column is $e_n/n^2.$ Then your $J= A -2B.$ Now $B = \lim_{N\to \infty} B_N$ in the operator norm, where $B_N$ is the matrix that agrees with $B$ in the first $N$ columns, and is $0$ in the other columns. Note that $A-2B_N$ is finite rank, having range equal to the span of $v,e_1,\dots, e_N.$ Since $\|(A-2B_N) - (A-2B)\|\to 0,$ $J= A-2B$ is compact as desired.
</s>