I am stuck at 2.8.7b. The author suggested the strategy in the proof of theorem 2.8.1. But I can't see how the strategy can be deployed. As attached, the proof of theorem 2.8.1 involves 2 variables whereas 2.8.7b involves only 1 variable.
I just don't know how. I can't see how $s_{nn}$ is related in this case. If I tried to minus $d_k$ from $s_{kk}$, it's just another triangle sum.
Since $t_{nn}$ converges, it is a Cauchy sequence. Hence, \begin{align} \forall\epsilon\in\mathbb{R}^+\exists n_1\in\mathbb{N}\forall n,m\geq n_1\rightarrow\lvert t_{nn}-t_{mm}\rvert<\epsilon \end{align} Hence, \begin{align*} \forall\epsilon\in\mathbb{R}^+\exists n_1\in\mathbb{N}\forall n\geq 2n_1,\\ \left\lvert\sum_{k=2}^{n} d_k-\sum_{i,j=1}^{n}a_{ij}\right\rvert&\leq\left\lvert\sum_{i,j=2}^{n}\lvert a_{ij}\rvert-\sum_{i,j=2}^{\left\lfloor\frac{n}{2}\right\rfloor}\lvert a_{ij}\rvert\right\rvert\text{ (see footnote)}\\ &=\left\lvert t_{nn}-t_{\left\lfloor\frac{n}{2}\right\rfloor\left\lfloor\frac{n}{2}\right\rfloor}\right\rvert\\ &<\epsilon \end{align*} Therefore, \begin{align} d_{kk}-s_{nn}\to 0\\ s_{nn}\to\mathbb{S} \end{align} By algebraic limit theorem, $d_{kk}\to$ S.
Footnote: This inequality can be visualized as such. $s_{nn}$ is the sum of a square sized with $n^2$ elements. $d_{kk}$ is the sum of elements in the upper left triangle. Hence, the difference of two will be the sum of elements in the lower right triangle. By triangle inequality, they are less than the sum of absolute terms. However, I can't come up with a Cauchy sequence that possesses this property. So I turn to $t_{nn}-t_{\left\lfloor\frac{n}{2}\right\rfloor\left\lfloor\frac{n}{2}\right\rfloor}$, which turns out cover slightly more than the lower right triangle.