Convergence of infinite matrices Im having difficulty proving the following result.
If $A$ is a matrix with countably infinite dimension, positive entries 
and $A_n$ is a sequence of matrices defined by
$$
(A_n)_{i,j} = \begin{cases}
(A)_{i,j} &\mbox{if } i,j \leq n\\
0 &\mbox{ else}
\end{cases}
.
$$
Then how can I prove that $A_n$ converges to $A$ in the $\ell^2$-operator norm?
 A: This is not true. Assuming you express your matrices with respect to an orthonormal basis, you can let $A=I$, i.e., 
$$
A_{kj}=\begin{cases} 1,&\ \text{ if } k=j\\ 0,&\ \text{ if } k\ne j\end{cases}.
$$
Then $\|A-A_n\|=1$ for all $n$, so the sequence $\{A_n\}$ does not converge to $A$. 
It does converge in the strong operator topology, though. Convergence in the strong operator topology is pointwise convergence of operators. The $A_n$ you defined are given by $A_n=P_nAP_n$, where $P_n$ is the orthogonal projection onto the subspace spanned by the first $n$ elements of the canonical basis. It is easy to check that 
$$
\|x-P_nx\|\to0
$$
for all $x\in\ell^2$, and so $P_n\to I$ in the strong operator topology. Then
\begin{align}
\|A_nx-Ax\|&=\|P_nAP_nAx-Ax\|\leq\|P_nA(P_n-I)x\|+\|(P_n-I)Ax\|\\ \ \\
&\leq\|(P_n-I)x\|+\|(P_n-I)Ax\|\to0.
\end{align}
This depends on $A$ being given by a bounded operator on $\ell^2$ (note that many matrices aren't, and it is not easy to determine in general which matrices are valid).
