Let's say I have a $symmetric$ matrix with components $\{ T_{jk} \}_{j,k=1}^{n}$ satisfying $T_{jk} = T_{kj}$.
Is there any way to simplify the following summation? $$ \sum_{k=1}^{n} \tfrac{1}{k} \sum_{j=1}^{k-1} T_{jk} $$
This is related to this one of my previous questions.
I have a hunch that I can get this into the form: $$ \alpha \sum_{j,k=1}^{n} T_{jk} - \beta \sum_{j=1}^{n} T_{jj} $$ for some constants $\alpha,\beta$. But I can't quite see how this would work. Is there a way to deal with such a summation?
No, there is no desired way to simplify this sum. It is pretty easy to see for $n = 3$: $$\sum_{k=1}^{3} \tfrac{1}{k} \sum_{j=1}^{k-1} T_{jk} = \frac12 T_{1, 2} + \frac13 T_{1, 3} + \frac13 T_{2, 3}.$$ And all these summands have the same coefficient $2\alpha$ in sum $$\alpha \sum_{j,k=1}^{n} T_{jk} - \beta \sum_{j=1}^{n} T_{jj}.$$ Also these three summands with the same coefficients present for all $n \ge 3$.
However it is easy to see that for $n \le 2$ we have $\alpha = \beta = \frac14$.