Is there a sequence $(t_n)_{n\ge 0}$ such that $\sup_n S_n < \infty$?

Question

Let $(t_n)_{n\ge 0}$ be a strictly increasing sequence of non-negative real numbers such that $t_0 =0$ and $\lim_n t_n = \infty$. We define an induced sequence $(S_n)_{n\ge 1}$ by $$ S_n := \sum_{i=0}^{n-1} \frac{t_{i+1} - t_i}{t_n - t_i}. $$

I would like to ask if there is a sequence $(t_n)_{n\ge 0}$ such that $\sup_n S_n < \infty$. Thank you so much for your elaboration!

Answer 1

Yes, you can take $t_n=2^n$ for $n \ge 1$. Then $$ \frac{t_{i+1} - t_i}{t_n - t_i} = \frac{2^{i+1}-2^i}{2^n-2^i} = \frac{2^i}{2^i(2^{n-i} -1)} = \frac{1}{2^{n-i}-1} \le \frac{1}{2^{n-i-1}} $$ and since the geometric series converges, $S_n$ is bounded

Answer 2

Let $a>1$, $t_n = a^n-1$, so $$S_n = \sum_{i=0}^{n-1} \frac{a-1}{a^{n-i} -1} < \frac{a}{a-1}$$