A question about the cyclic property of traces for infinite dimensional matrices Does the cyclic property of trace hold for infinite dimensional matrices (which are often encountered in quantum mechanics)? In other words, does the identity $$\rm Tr(ABC)=Tr(CAB)=Tr(BCA)$$ hold when all the matrices $A, B, C$ are infinite dimensional? 
 A: An example with two matrices (it can be made three by adding the identity matrix to the problem) indicates for me, that the traces can be different. In this example both traces lead to divergent infinite series, but the scheme of the matrix-contructions might allow simple variations which lead to convergent traces (see second example in the final conclusion).           
Preliminaries and construction of matrices


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*Assume the infinite sized Pascal-matrix $P$ (upper triangular) 
$ P[r,c]= \binom cr \quad$ (the row- and column-indices r and c begin at 0).     

*Next assume a rowvector  $V(x)=[1,x,x^2,x^3,...]$ again of infinite size.              
For the dot-product we would get (simply by the binomial theorem)
$$ V(x) \cdot P = V(x+1) \tag 1$$
but also $V(x) \cdot P^2= V(x+2)$ and $V(x) \cdot P^{-1}=V(x-1)$ and so on.   

*Next assume the infinite-sized matrix $S_2$ (lower triangular) containing the Stirlingnumbers 2nd kind, similarity-scaled by vector and reciprocal vector of factorials $$S_2[r,c]=s_2(r,c)\cdot c!/r!$$  such that with a vector $V(x)$ we would get
$$V(x) \cdot S_2 = [1, e^x-1, (e^x-1)^2, (e^x-1)^3,...] = V(e^x-1) \tag 2$$
(That identity is well known, see for instance the HoMF (Abramowitz/Stegun))

*Of course, appending a multiplication with $P$ gives a nicer result
$$V(x) \cdot S_2 \cdot P = V(e^x-1) \cdot P = V((e^x-1)+1)=V(e^x) \tag 3$$
Doing the dotproduct $S_2 \cdot P$ first we get a very nice looking matrix $B$:
$$ B = S_2 \cdot P \qquad \qquad \text{ where } B[r,c]= {c^r \over r!} $$
The matrix has a diagonal with a very simple form $c^c/c!$ of which we shall take advantage now.

Now we discuss the traces of compositions $P^{-1} \cdot B$ and $B \cdot P^{-1}$.  To make this triple products we could insert the identity matrix here, but leave this aside.
1) The trace of $P^{-1} \cdot B$
It is not obvious what the entries of $P^{-1} \cdot B$ look like, but we can derive this by a trick.
We consider the matrix-product extended by the leftmultiplication with the indeterminate $V(x)$ and use associativity
$$ \begin{array} {} V(x) \cdot P^{-1} \cdot B &= (V(x) \cdot P^{-1}) \cdot B\\
 &= V(x-1) \cdot B \\
 &= V(e^{x-1}) \end{array} \tag 4$$
But the latter can be rewritten as
$$ V(e^{x-1})= V(e^x/e) = V(e^x) \cdot \;^dV(1/e)  \tag 5$$
(where the little prefix "d" means that the vector is used as diagonalmatrix)
$\qquad \qquad$ and this means we have also
$$ \begin{array} {} 
V(x) \cdot P^{-1} \cdot B &=V(x) \cdot B \cdot \;^dV(1/e) \\
&=V(x) \cdot (B \cdot \;^dV(1/e)) \\
&= V(x) \cdot A \end{array} \tag 6 $$
and the right-hand dot-product $A$ means a column-scaling of the $B$-matrix such that the diagonal elements are 
$$a[c,c]={c^c\over c!} \cdot {1 \over e^c} ={(c/e)^c\over c!}$$
The trace of $P^{-1} \cdot B = A $ is now formally the infinite and divergent sum $$ \text{TR}(A) = \sum_{k=0}^\infty {(c/e)^c \over c!} \tag 7 $$which reduces to an approximation to a sort of geometric series and might have some analytic continuation.
Added: Using W|A I found the generating function for this series, when the $e$ in the summands is taken as formal indeterminate $x$ :
$$ \sum_{k=0}^\infty {(c x)^c \over c!} =  {1 \over W(-x)+1} \tag {7.1} $$
which is only convergent if $x< \frac1e$, so for instance if $x= \frac 1{e^2}$ and thus if we use $P^{-2}$ instead of $P^{-1}$ in (6).      
2) The trace of $ B \cdot P^{-1} $
The other trace is trivial to determine. We get
$$\begin{array} {} \text{TR} (B \cdot P^{-1}) &= \text{TR} ((S_2 \cdot P) \cdot P^{-1}) \\
&= \text{TR} (S_2 \cdot( P \cdot P^{-1})) \\
&= \text{TR} (S_2)  \\
&= \displaystyle \sum_{k=0}^\infty (1 )
\end{array} \tag 8$$
3) possible conclusion and improvement of the argument
Those results are of course formally completely different divergent series and possibly this whole argument suffices to show that the infinite matrices do not always inherit the cyclicity from the finite matrix.              
To improve the example/argument I looked at the traces of
$P^{-2} \cdot B = A_2 $ and $B \cdot P^{-2} = S_2 \cdot P^{-1} =T_2 $ and while I got a convergent series now for the trace of $A_2$ evaluating to  $\text{TR}(A_2) = {1 \over 1+ W(-1/e^2)} \approx 1.18848736943$ there occurs in $\text{TR}(T_2)$ some series with irregularly alternating signs, where the absolute values of the terms $t[c,c]$ also diverge when I look at $64$ or $128$ terms and which seems not to be Cesaro- or Eulersummable. 
