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victorp
Pe: 2017-01-06T17:45:48+02:00 In: In:

Limita Sir

Sirul de numere pozitive (an) n>=1 satisface
\lim_{n \to \infty} (\dfrac{a1}{a2} +\dfrac{a2}{a3} +...+\dfrac{a(n-1)}{an}+\dfrac{an}{a1}-n)=0
Sa se arate ca sirul an este constant.
Si stie cineva unde pot gasi probleme si rezolvari din GM din anii trecuti?

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3 raspunsuri

  1. DD
    profesor

    Fie P(n)=>lim(n=>∞)[a1/a2+a2/a3++a(n-1)/an+an/a1-n]=0,pentru orice ak>0, atunci sirul an este constant,adica a1=a2=..=an
    1) P(2)->a1/a2+a2/a1-2=0=>a1/a2=1 sau a1=a2 de unde P(2) adevarat
    2) Fie P(k) adevara,adica pentru; a1/a2+a2/a3+..+a(K-1)/ak+ak/a1-k=0–>=>a1=a2=.=ak
    3) Sa se arate ca si P(K+1) este adevarat. Deci a1/a2a2/a3+..+a(k-1)/aK+ak/a(k+1)+ak+1)/a1-(k+1)=0 sau P(k+1)-P(k)–>=>ak/a(k+1)+a(k+1)/a1-(k+1)-ak/a1+k=0 sau;(a(k+1)-ak)/a1+ ak/a(k+1)-1=0sau; (a(k+1)-ak)/a1=(a(k+1)-ak)/a(k+1),sau a1=a(k+1) sau a1=a2=..ak=a(k+1) deci adevarat.de unde si P(n) adevarat

  2. gigelmarga
    profesor

    DD wrote: Fie P(n)=>lim(n=>∞)[a1/a2+a2/a3++a(n-1)/an+an/a1-n]=0,pentru orice ak>0, atunci sirul an este constant,adica a1=a2=..=an
    1) P(2)->a1/a2+a2/a1-2=0=>a1/a2=1 sau a1=a2 de unde P(2) adevarat
    2) Fie P(k) adevara,adica pentru; a1/a2+a2/a3+..+a(K-1)/ak+ak/a1-k=0–>=>a1=a2=.=ak
    3) Sa se arate ca si P(K+1) este adevarat. Deci a1/a2a2/a3+..+a(k-1)/aK+ak/a(k+1)+ak+1)/a1-(k+1)=0 sau P(k+1)-P(k)–>=>ak/a(k+1)+a(k+1)/a1-(k+1)-ak/a1+k=0 sau;(a(k+1)-ak)/a1+ ak/a(k+1)-1=0sau; (a(k+1)-ak)/a1=(a(k+1)-ak)/a(k+1),sau a1=a(k+1) sau a1=a2=..ak=a(k+1) deci adevarat.de unde si P(n) adevarat

    E o glumă, sau cum?

  4. victorp

    Am ajuns la
    \lim_{n \to \infty}(a(n-1)-an)=0 \\ sau \\ \lim_{n \to \infty}(an)=a1;
    Daca ajuta
    Fie sirul
    bn = \frac{a1}{a2}+\frac{a2}{a3}+...+\frac{a(n-2)}{a(n-1)}+\frac{a(n-1)}{an}+\frac{an}{a1}
    Deci
    b(n-1)=\frac{a1}{a2}+\frac{a2}{a3}+...\frac{a(n-2)}{a(n-1)}+\frac{a(n-1)}{a1}
    Prin scadere se obtine
    bn=b(n-1)+\frac{a(n-1)}{an}+\frac{an}{a1}-\frac{a(n-1)}{a1}
    \lim_{n \to \infty}(bn-n)=0
    Deci
    \lim_{n \to \infty}(b(n-1)-(n-1)=0 \\ \lim_{n \to \infty}(bn-n)=\lim_{n \to \infty}(b(n-1)-(n-1) + \frac{a(n-1)}{an}+\frac{an-a(n-1)}{a1}-1)=\lim_{n \to \infty}(b(n-1)-(n-1) + \frac{a(n-1)-an}{an}+\frac{an-a(n-1)}{a1}) \\ Cum \lim_{n \to \infty}(b(n-1)-(n-1)=0 \\ lim_{n \to \infty}(bn-n)=lim_{n \to \infty}( \frac{a(n-1)-an}{an}+\frac{an-a(n-1)}{a1})=0 =>lim_{n \to \infty}[(a(n-1)-an)\frac{1}{an}]=\lim_{n \to \infty}[(a(n-1)-an)\frac{1}{a1}]
    Ceea ce inseamna ori ca
    \lim_{n \to \infty}(a(n-1)-an)
    ori ca
    \lim_{n \to \infty}(an-a1)=0

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