Question

va rog daca ma puteti ajuta cat mai repede :)

Se considera sirul $I_{n}$ n≥1, unde $I_{n}$ = $\int\limits^1_0 { \frac{ x^{n} }{ x^{2} + x + 1} } \, dx$

Demonstrati ca $\frac{1}{3(n+1)}$≤ $I_{n}$≤ $\frac{1}{3(n-1)}$ si calculati lim n-> ∞ $I_{n}$

Answer 1

Aratam ca sirul este descrescator.
[tex]I_{n+1}-I_n= \int\limits^1_0 { \frac{x^n(x-1)}{x^2+x+1} } \, dx \leq 0~deoarece\\ x-1 \leq 0,~pentru~x\in[0,1]\\ x^n \geq 0~pentru~x\in[0,1]\\ x^2+x+1\ \textgreater \ 0pentru~x\in R\\ I_{n+2}+I_{n+1}+I_n= \int\limits^1_0 {x^n} \, dx = \frac{1}{n+1} \\ Din:\\ I_{n+2} \leq I_n,I_{n+1} \leq I_n,I_{n+} \leq I_n~rezulta~prin~insumare:\\ \frac{1}{n+1} \leq 3I_n=\ \textgreater \ \frac{1}{3(n+1)} \leq I_n [/tex]
[tex]Din:\\ I_{n+2} \leq I_n,I_{n+2} \leq I_{n+1},\\ I_{n+2} \leq I_{n+2} \\ rezulta~prin~insumare:\\ I_{n+2} \leq \frac{1}{3(n+1)} \\ si~inlocuind~n~cu~n-2~obtinem:\\ I_n} \leq \frac{1}{3(n-1)} \\ In~final:\\ \frac{1}{3(n+1)} \leq I_n} \leq \frac{1}{3(n-1)} ~si~trecand~la ~limita\\ conform~criteriului~clestelui\\ \lim_{n \to \infty} I_n = \frac{1}{3} [/tex]