Do these series converge to logarithms?

You can write $T(n,k)=1-n1_{\{n\mid k\}}$. Then, for $\vert x\vert < 1$ look at the power series $$ \begin{align} \sum_{k=1}^\infty\frac{T(n,k)}{k}x^k&=\sum_{k=1}^\infty\frac{x^k}{k}-\sum_{k=1}^\infty1_{\{n\mid k\}}\frac{nx^k}{k}\\ &=\sum_{k=1}^\infty\frac{x^k}{k}-\sum_{k=1}^\infty\frac{x^{nk}}{k}\\ &=-\log(1-x)+\log(1-x^n)\\ &=\log\left(\frac{1-x^n}{1-x}\right)\\ &=\log(1+x+\cdots+x^{n-1}). \end{align}. $$ So, letting $x$ increase to 1, $$ \lim_{x\uparrow1}\sum_{k=1}^\infty\frac{T(n,k)}{k}x^k=\log n. $$ The fact that you can commute this limit with the summation to get $\sum_{k=1}^\infty T(n,k)/k$ follows from the fact the series converges uniformly (over $0 < x < 1$). You can show this by grouping together the consecutive positive terms where $n\nmid k$ to get a sequence with alternating signs and decreasing in magnitude. Then, truncating the series gives an error which is bounded by the following term. That is, $$ \left\vert\sum_{k=1}^{jn-1}\frac{T(n,k)}{k}x^k-\log(1+x+\cdots+x^{n-1})\right\vert \le \frac{-T(n,jn)}{jn}x^{jn}\le \frac1j. $$ Commuting the limit with a finite sum is no problem, so you get $$ \left\vert\sum_{k=1}^{jn-1}\frac{T(n,k)}{k}-\log n\right\vert\le\frac1j. $$

Yes. You can get the sums by differentiating the digamma function repeatedly. There is a good deal of information about the resulting polygamma functions, including series expressions, here. Your matrix version is a lot more visually arresting than the usual Dirac delta function formulation!