Romanian District Olympiad 2019 - Grade 11 - Problem 1

Art of Problem Solving

AoPS Online

College Math Topics in undergraduate and graduate studies

Topics in undergraduate and graduate studies

3 M G

BBookmark VNew Topic kLocked

College Math Topics in undergraduate and graduate studies

Topics in undergraduate and graduate studies

3 M G

BBookmark VNew Topic kLocked

w

4 users

TOPICS

No tags match your search

M

No tags match your search

M

Romanian District Olympiad 2019 - Grade 11 - Problem 1

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: Romanian District Olympiad 2019 - Grade 11 - Problem 1

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

867 posts

#1 h Mar 16, 2019, 8:50 PM • 3 Y

Y by Mathuzb, Adventure10, Mango247

Let $(a_n)_{n \ge 1}$ be a sequence of positive real numbers such that the sequence $(a_{n+1}-a_n)_{n \ge 1}$ is convergent to a non-zero real number. Evaluate the limit $\lim_{n \to \infty} \left( \frac{a_{n+1}}{a_n} \right)^n.$

This post has been edited 1 time. Last edited by Filipjack, Mar 16, 2019, 8:51 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

407 posts

DerJan

#2 h Mar 16, 2019, 10:22 PM • 3 Y

Y by Mathuzb, Adventure10, Mango247

Since $(a_n)$ is a sequence of positive reals $a=\lim_{n \to \infty} a_{n+1}-a_n$ is positive. For each $1>\epsilon>0$ there is a natural number $N$ such that for all $n\geqslant N$
$\begin{align*} (1-\epsilon)an&<a_n<(1+\epsilon)an \text{ and} \\ (1-\epsilon)a&<a_{n+1}-a_n<(1+\epsilon)a \\ \Rightarrow 1+\frac{1-\epsilon}{(1+\epsilon)n}&<\frac{a_{n+1}}{a_n}<1+\frac{1+\epsilon}{(1-\epsilon)n}. \end{align*}$ By squeeze theorem we get
$\exp\left(\frac{1-\epsilon}{1+\epsilon}\right)<\lim_{n \to \infty} \left(\frac{a_{n+1}}{a_n}\right)^n<\exp\left(\frac{1+\epsilon}{1-\epsilon}\right),$ so $\lim_{n \to \infty} \left(\frac{a_{n+1}}{a_n}\right)^n=e$ .

This post has been edited 1 time. Last edited by DerJan, Mar 16, 2019, 10:28 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1512 posts

mathuz

#3 h Mar 16, 2019, 11:58 PM • 2 Y

Y by Adventure10, Alphaamss

From the Stolz-Cesaro, we get that $\lim_{n \rightarrow \infty} \frac{a_n}{n} = \lim_{n\rightarrow \infty} (a_{n+1} - a_{n}) = l \neq 0.$ So $(a_n)$ is asymptotically same with $(n)$

.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

206 posts

#4 h Jun 7, 2019, 11:30 PM • 2 Y

Y by Adventure10, Mango247

Filipjack wrote:

Let $(a_n)_{n \ge 1}$ be a sequence of positive real numbers such that the sequence $(a_{n+1}-a_n)_{n \ge 1}$ is convergent to a non-zero real number. Evaluate the limit $\lim_{n \to \infty} \left( \frac{a_{n+1}}{a_n} \right)^n.$

It was proposed by me in the Mathematical Gazette.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

207 posts

#5 h Apr 5, 2025, 7:06 AM

Y by

Pencil nib sacrifice

Attachments:

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a