Let $T$ be an operator on an infinite-dimensional Hilbert space $H$, $r$ be its spectral radius. Let $\phi$ be a positive linear functional on the algebra $B(H)$ of bounded operators on $H$ such that $\phi(I_H)=1$ and $\phi(S^*S)=0$ implies $S=0$. I wonder if $$\sum_{n=1}^\infty\dfrac{\phi(T^n)}{nr^n}$$ diverges?
Informally we can have $\phi(\log|T-rI|)$ which seems to diverge, but how should I prove it formally?
Jonas provided a nice counterexmaple where the series converges. But I think it is because he constructed an operator which has $i,-i$ as its spectrum, then the series is reduced to $\sum i^n/n$ and $\sum (-i)^n/n$ which both converge.
I wonder what if I assume $T$ has an eigenvalue which equals to $r$?
Jonas gave another counterexmaple. Please see the answer below. I admit that my intuition is for when $\phi$ is a trace and $T$ is a matrix, then $\phi(T^n)$ will be $\sum_i \lambda_i^n$, where $\lambda_i$ are eigenvalues. Then it will diverge. But what about a general operator with a general tracial positive linear functional?
This part can be found here
Let $(e_n)$ be an orthonormal basis for $H$, and let $\phi$ be defined by $\phi(A)=\sum\limits_{k=1}^\infty 2^{-k}\langle Ae_k,e_k\rangle$. Let $T$ be the bounded operator such that $Te_1=e_2$, $Te_2=-e_1$, and $Te_k=0$ if $k>2$. Then $r=1$, and $\phi(T^n)$ is $0$ when $n$ is odd, $-\frac34$ when $n$ is even and not a multiple of $4$, and $\frac34$ when $n$ is a multiple of $4$. The series will converge by the alternating series test.
For the additional question, how about on $M_2$ this time, leaving out the details of extending the idea to $B(H)$, $\phi\left(\begin{bmatrix}a&b\\c&d\end{bmatrix}\right)=\frac12(a+d)-\frac14(b+c)$, and $T=\begin{bmatrix}1&2\\0&0\end{bmatrix}$.