Question

Sa se calculeze limita sirului $x_{n}=\prod\limits^n_{k=2} \frac{k^2+k-2}{k(k+1)}$. Mersi

Answer 1

$x_n=\prod_{k=2}^n\left[\dfrac{k^2+k-2}{(k+1)\cdot k}\right]=\prod_{k=2}^n\left[\dfrac{k^2-1+k-1}{(k+1)\cdot k}\right]=\\=\prod_{k=2}^n\left[\dfrac{(k-1)\cdot(k+1)+k-1}{(k+1)\cdot k}\right]=\prod_{k=2}^n\left[\dfrac{(k-1)\cdot(k+2)}{(k+1)\cdot k}\right]=\\=\left(\dfrac{1}{3}\cdot\dfrac{4}{2}\right)\cdot\left(\dfrac{2}{4}\cdot\dfrac{5}{3}\right)\cdot\left(\dfrac{3}{5}\cdot\dfrac{6}{4}\right)\cdot\ldots\cdot\left(\dfrac{n-2}{n}\cdot\dfrac{n+1}{n-1}\right)\cdot\left(\dfrac{n-1}{n+1}\cdot\dfrac{n+2}{n}\right)=\\=\dfrac{1}{3}\cdot\dfrac{n+2}{n}=\dfrac{n+2}{3n}\to\dfrac{1}3.$

Este cumva problema AL 96 din culegerea de admitere la Politehnica din Timișoara ? Enunțul nu este chiar cel din culegere, dar acolo se ajunge după explicitarea combinărilor :-).

Green eyes.

Answer 2

= produs dupa k de la 2 la n din
(k^2  - 1  + k  - 1) / (k(k+1)) = (k-1)(k+2) : (k(k+1)) =
1x4x2x5x3x6x...x(k-1)(k+2) totul supra 2x3x3x4x4x5x...k(k+1)=
(n+2) / 3n si astfel avem limita cand n tinde la infinit din produsul nostru =
1/3