$\limsup_{n\to\infty} \left(s_{n} / t_{n}\right) \le \limsup_{n\to\infty} \left( \left(s_{n+1}-s_{n}\right) / ( \left(t_{n+1}-t_{n}\right)\right)$

Question

I am trying to solve the following problem.

Problem Suppose that $\left( s_{n} \right)_{n=0}^{\infty}$ is a sequence of real numbers and that $\left( t_{n} \right)_{n=0}^{\infty}$ is a strictly increasing unbounded sequence of positive numbers. Show that

$$\limsup_{n\to\infty} \left(s_{n} / t_{n}\right) \le \limsup_{n\to\infty} \left( \left(s_{n+1}-s_{n}\right) / ( \left(t_{n+1}-t_{n}\right)\right).$$

I approached it by splitting the sequence $\left( \sup \{ s_{k}/t_{k} \;| \; k \ge n \} \right)_{n=0}^{\infty}$ into two cases, one where it is bounded below and one where it is not bounded below, because the sequence $\left( \sup \{ s_{k}/t_{k} \;| \; k \ge n \} \right)_{n=0}^{\infty}$ is decreasing. But after that I couldn't figure out how to solve it. Even the smallest hint would be very helpful.

Answer 1

I'll show that if $$ \limsup_{n\to\infty} \frac{s_{n+1}-s_{n}}{t_{n+1}-t_{n}}<a\qquad (*) $$ then $$ \limsup_{n\to\infty} \frac{s_{n}}{t_{n}}<a $$ which is sufficient for your claim. Indeed, under (*) there exists an $N>0$ such that for all $i\ge N$ $$ s_{i+1}-s_{i}\le (t_{i+1}-t_{i})a. $$ Fix $n>N$ and sum up this inequality for $i=N,N+1,...,n-1$ and get $$ s_{n}-s_{N}< (t_{n}-t_{N})a $$ yielding $$ s_{n}< s_{N}+ (t_{n}-t_{N})a \Longrightarrow \frac{s_n}{t_n}< a+\frac{s_{N}}{t_n}. $$ Since $t_n\to+\infty$, eventually $s_n/t_n<a$ as required.