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PallMall
Pe: 2013-10-25T16:04:44+03:00 In: In:

Limita admitere politehnica

ati putea careva sa ma ajutati … macar sa scriu suma altfel. Ca de ceva timp imi bat capu si nu imi vine in minte sa o scriu altfel sa se reduca termenii

    \[ {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{{k(k + 1)}}{{2x^{k - 1} }} = ?\,\,\,x > 0} \]

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4 raspunsuri

  1. sandy_sc
    maestru (V)

    Pt x=1 obtii o suma de sume gauss a caror llimita este +oo

    ex
    S=1+(1+2)+(1+2+3)+…+(1+2+…+n) SUMA CARE EVIDENT TINDE LA +OO

    daca x apartine (0, 1) OBTINEM O sua S` .Comparand termenii sumei S` cu cei ai lui S oservam ca termeni sumwei S` sunt mai mari decat ai lui S, putem spune ca lim S`__>+OO

    Cazul x>1 mai ma gandesc

  2. sandy_sc
    maestru (V)

    pT X>1 SE aplica Criteril General a lui Cauchy

    S(n+p)=1+2*3/2*X+3*4/2*X^2+…+n*(n+!)/2*x^(n-1)+(n+1)*(n+2)/2*X^n+…+(n+p)*(n+p+!)/2*x^(n+p-1)

    S(n)=1+2*3/3*X+3*4/2X^2+…+n*(n+!)/2*x^n-1

    lSn+p_Snl=l(n+1)*(n+2)/2*x^n+…+(np)*)n+p+1)/2*x^(n+p+1)l

    nOTAm cu e (epsilon) =1/x^n+1/x^*(n+1)+…+1/x^(n+p+1)
    se obseva ca termenii lui e sunt in progresie geoetrica a caror suma este
    S(e)=1/X^(n+!)*[1-(X)^(n+p+2)]/(1-X)
    ….

    CONF cRIT LUI cAUCHY SIRUL E DIVERGENT , ARE LIMITA LA +OO

  4. ghioknt
    profesor

    Preiau de la sandy_sc: k(k+1)/2=1+2+3+…+k.

        \[ \begin{array}{l} a_n = 1 + \frac{{1 + 2}}{x} + \frac{{1 + 2 + 3}}{{x^2 }} + \frac{{1 + 2 + 3 + 4}}{{x^3 }} + ... + \frac{{1 + 2 + ... + \left( {n - 2} \right)}}{{x^{n - 3} }} + \frac{{1 + 2 + ... + \left( {n - 1} \right)}}{{x^{n - 2} }} + \frac{{1 + 2 + ... + n}}{{x^{n - 1} }} \\ \frac{{a_n }}{x} = \frac{1}{x} + \frac{{1 + 2}}{{x^2 }} + \frac{{1 + 2 + 3}}{{x^3 }} + \frac{{1 + 2 + 3 + 4}}{{x^4 }} + ... + \frac{{1 + 2 + ... + \left( {n - 2} \right)}}{{x^{n - 2} }} + \frac{{1 + 2 + ... + \left( {n - 1} \right)}}{{x^{n - 1} }} + \frac{{1 + 2 + ... + n}}{{x^n }} \\ a_n - \frac{{a_n }}{x} = 1 + \frac{2}{x} + \frac{3}{{x^2 }} + \frac{4}{{x^3 }} + ... + \frac{{n - 1}}{{x^{n - 2} }} + \frac{n}{{x^{n - 1} }} - \frac{{1 + 2 + ... + n}}{{x^n }} \\ a_n \left( {1 - \frac{1}{x}} \right) \cdot \frac{1}{x} = \frac{1}{x} + \frac{2}{{x^2 }} + \frac{3}{{x^3 }} + \frac{4}{{x^4 }} + ... + \frac{{n - 1}}{{x^{n - 1} }} + \frac{n}{{x^n }} - \frac{{1 + 2 + ... + n}}{{x^{n + 1} }} \\ a_n \left( {1 - \frac{1}{x}} \right) - a_n \left( {1 - \frac{1}{x}} \right) \cdot \frac{1}{x} = 1 + \frac{1}{x} + \frac{1}{{x^2 }} + ... + \frac{1}{{x^{n - 1} }} - \frac{{n\left( {n + 1} \right)}}{{2x^n }} - \frac{n}{{x^n }} + \frac{{n\left( {n + 1} \right)}}{{2x^{n + 1} }} \\ a_n \left( {1 - \frac{1}{x}} \right)\left( {1 - \frac{1}{x}} \right) = \frac{{1 - \frac{1}{{x^n }}}}{{1 - \frac{1}{x}}} - \frac{{n\left( {n + 3} \right)}}{{2x^n }} + \frac{{n\left( {n + 1} \right)}}{{2x^{n + 1} }} \Rightarrow a_n = \frac{{1 - \frac{1}{{x^n }}}}{{\left( {1 - \frac{1}{x}} \right)^3 }} - \frac{1}{{\left( {1 - \frac{1}{x}} \right)^2 }}\left[ {\frac{{n\left( {n + 3} \right)}}{{2x^n }} - \frac{{n\left( {n + 1} \right)}}{{2x^{n + 1} }}} \right] \\ \lim a_n = \frac{1}{{\left( {1 - \frac{1}{x}} \right)^3 }} = \left( {\frac{x}{{x - 1}}} \right)^3 \quad deoarece\quad \lim \frac{{n^p }}{{x^n }} = 0,\quad daca\;x > 1,\quad chiar\;daca\;p > 0. \\ \end{array} \] \[ \begin{array}{l} \]

    Cu bine, ghioknt.

  5. PallMall

    mersi mult …nu observasem o chestie (cam evidenta) ..

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