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Se considera sirul \( a_n=\sqrt[3]{n^3+3n^2+2n+1}+a\sqrt[5]{n^5+5n^4+1}+\frac{\ln(e^{n^2}+n+2 )}{n+2}+b. \)

Sa se determine \( a,b \in \mathbb{R} \) astfel incat \( \lim\limits_{n \to \infty}a_n=5 \).

Concursul interjudetean "Teodor Topan", Subiectul III

\( a_n=(\sqrt[3]{n^3+3n^2+2n+1}-n)+(a\sqrt[5]{n^5+5n^4+1}-an)+b+(\frac{\ln(e^{n^2}+n+2)}{n+2}+n(a+1)) \)

\( \lim\limits_{n\to\infty}(\sqrt[3]{n^3+3n^2+2n+1}-n)=\lim\limits_{n\to\infty}\frac{n^3+3n^2+2n+1-n^3}{(\sqrt[3]{n^3+3n^2+2n+1})^2+n\sqrt[3]{n^3+3n^2+2n+1}+n^2}=\lim\limits_{n\to\infty}\frac{3+\frac{2}{n}+\frac{1}{n^2}}{(\sqrt[3]{1+\frac{3}{n}+\frac{2}{n^2}+\frac{1}{n^3}})^2+\sqrt[3]{1+\frac{3}{n}+\frac{2}{n^2}+\frac{1}{n^3}}+1}=1 \)
Analog aratam (folosind conjugata) ca \( \lim\limits_{n\to\infty}(a\sqrt[5]{n^5+5n^4+1}-an)=a \)
Avem deci ca \( \lim\limits_{n\to\infty}a_n=1+a+b+\lim\limits_{n\to\infty}(\frac{\ln(e^{n^2}+n+2)}{n+2}+n(a+1)) \)
\( \lim\limits_{n\to\infty}(\frac{\ln(e^{n^2}+n+2)}{n+2}+n(a+1))=\lim\limits_{n\to\infty}(\frac{\ln e^{n^2}+\ln(1+\frac{n+2}{e^{n^2}})}{n+2}+n(a+1))=\lim\limits_{n\to\infty}(\frac{n^2}{n+2}+n(a+1)+\ln(1+\frac{n+2}{e^{n^2}})^{\frac{1}{n+2}}) \)
Avem ca \( \lim\limits_{n\to\infty}(\ln(1+\frac{n+2}{e^{n^2}})^{\frac{1}{n+2}})=\lim\limits_{n\to\infty}(\ln((1+\frac{n+2}{e^{n^2}})^{\frac{e^{n^2}}{n+2}})^{\frac{1}{e^{n^2}}}))=ln (e^0)=0 \)
Ramane deci ca \( \lim\limits_{n\to\infty}a_n=1+a+b+\lim\limits_{n\to\infty}(n(a+1)+\frac{n^2}{n+2})=1+a+b+\lim\limits_{n\to\infty}(n(a+2)-2+\frac{4}{n+2}) \)
Pentru ca aceasta limita sa fie finita trebuie sa avem a=-2.
Avem astfel ca \( \lim\limits_{n\to\infty}a_n=b-1-2=5 \) Rezulta \( b=8 \).