Problema 4, lista scurta 2009

Moderators: Laurian Filip, Beniamin Bogosel, Filip Chindea

Post Reply
alex2008
Leibniz
Posts: 464
Joined: Sun Oct 19, 2008 3:23 pm
Location: Tulcea

Problema 4, lista scurta 2009

Post by alex2008 »

Se considera un sir \( (a_n)_{n\ge 1} \) de numere reale definit prin \( a_{n+1}=\frac{(1+a_n)}{(1+a_n^2)} \), pentru orice \( n\in \mathbb{N}^* \). Aratati ca daca \( a_1\in (0,2) \), atunci \( |a_{n+1}-1|\le \frac{1}{2^n} \), pentru orice \( n\in \mathbb{N}^* \).

Lucian Dragomir, Otelu-Rosu
. A snake that slithers on the ground can only dream of flying through the air.
Marius Mainea
Gauss
Posts: 1077
Joined: Mon May 26, 2008 2:12 pm
Location: Gaesti (Dambovita)

Post by Marius Mainea »

Inductie dupa n:

1) verificare, trivial.

2) \( |{a_{n+1}-1|=|\frac{a_n-a_n^2}{1+a_n^2}|\le\frac{a_n}{1+a_n^2}\cdot\frac{1}{2^{n-1}}\le \frac{1}{2^n} \)
Post Reply

Return to “Clasa a IX-a”