Show that $\limsup_{n\to\infty}\frac{x_{n+1}-x_1}n=\limsup_{n\to\infty}\frac{x_n}n$

Question

Let $(x_n)_{n\in\mathbb N}\subseteq[-\infty,\infty)$. What's the easiest way to show that Show that $\limsup_{n\to\infty}\frac{x_{n+1}-x_1}n=\limsup_{n\to\infty}\frac{x_n}n$?

Clearly, since $x_1/n\to0$, $\limsup_{n\to\infty}\frac{x_{n+1}-x_1}n=\limsup_{n\to\infty}\frac{x_{n+1}}n$, but how do we see that the latter is equal to $\limsup_{n\to\infty}\frac{x_n}n=\inf_{n\in\mathbb N}\sup_{k\ge n}\frac{x_n}n$?

Simplify the taks by generalizing: If $a_n\to 1$ and $b_n\to 0$, show $\limsup (a_nz_n+b_n)=\limsup z_n$ — Hagen von Eitzen, Apr 27 '20 at 15:26

score 1 · Accepted Answer · answered Apr 27 '20 at 15:27

1

Note that $\lim\sup \frac{x_{n+1}}{n}=\lim\sup \frac{n+1}{n}\frac{x_{n+1}}{n+1}$. The first fraction of the RHS obviously goes to 1.

Then use this: limsup of the product of two sequences, of which one converges.

answered Apr 27 '20 at 15:27

Dasherman

4,206

Show that $\limsup_{n\to\infty}\frac{x_{n+1}-x_1}n=\limsup_{n\to\infty}\frac{x_n}n$

1 Answers1