Given a natural number $n \ge 2$, we define
- a finite sequence $(t_{n,i})_{i=0}^n$ by $t_{n, 0} := 0, t_{n, n} := \frac{\pi^2}{6}$ and $$ t_{n, i+1} := t_{n, i} + \frac{1}{ (n - i)^2} \quad \forall i \in \{0, \ldots,n-2\}. $$
- a finite sequence $(S_{n,k})_{k=1}^{n}$ by $$ S_{n, k} :=\sum_{i=0}^{k-1} \frac{t_{n, i+1} - t_{n, i}}{t_{n, k} - t_{n, i}} \quad \forall k \in \{1, \ldots,n\}. $$
Could you elaborate if $$ \alpha :=\sup_{n \ge 2} \sup_{1 \le k \le n} S_{n, k} $$ is finite or not?
My attempt We have $$ \begin{align*} S_{n,k} &= 1+ \sum_{i=0}^{k-2} \frac{t_{n, i+1} - t_{n, i}}{\sum_{j=i}^{k-1} t_{n, j+1} - t_{n, j}} \\ &= 1 + \sum_{i=0}^{k-2} \frac{1}{(n-i)^2 \sum_{j=i}^{k-1} \frac{1}{(n-j)^2}}. \end{align*} $$
Then $$ \begin{align*} S_{n, n} &= 1 + \sum_{i=0}^{n-2} \frac{1}{(n-i)^2 \sum_{j=i}^{n-1} \frac{1}{(n-j)^2}} \\ &\le 1 + \sum_{i=0}^{n-2} \frac{1}{(n-i)^2} \\ &\le 1+ \frac{\pi^2}{6}. \end{align*} $$
Than I'm stuck at upper bounding $\sup_{1 \le k \le n-1} S_{n, k}$.
$$S_{n, k} = 1 + \sum_{\ell = n-k-2}^{n} \frac1{\ell^2} \times \frac1{\sum_{j=n-k-2}^\ell\frac1{j^2}}$$
Let $m=n-k-2$, then
\begin{align} S_{n,k} &= 1 + \sum_{\ell = m}^{n} \frac1{\ell^2}\times\frac1{\sum_{j=m}^\ell \frac1{j^2}}\\ &\ge \sum_{\ell = m}^n \frac1{\ell^2} \frac1{\displaystyle\int_{m-1}^{\ell}\frac1{x^2}\mathrm d x}\\ &= \sum_{\ell = m}^n \frac{m-1}{\ell(\ell - m + 1)}\\ &= \sum_{\ell=m}^n \frac{1}{\ell - m + 1} - \sum_{\ell=m}^{n}\frac1\ell\\ &= H_{n-m+1} - \left(H_n - H_{m-1}\right) \end{align}
This proves that $$S_{2n, n-1} \ge H_{n} - H_{2n} + H_{n-2} \to \infty$$
since $\ln n - \ln (2n) + \ln (n-2) = \ln(n-2) - \ln (2)\to \infty$
This proves that $\alpha = \infty$.