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Adrianx
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Pe: 2009-02-23T16:30:02+02:00 In: In:

Limite

Sa se calculeze limita
{\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^\infty {\log _{\frac{1}{2}} } \frac{{k(k + 2)}}{{(k + 1)^2 }} \]
Cum se calculeza genul acesta de limite? (sume Riemann sau ce???) Mai folositoare mi-ar fi niste explicatii decat o rezolvare. Multumesc anticipat!

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4 raspunsuri

  1. profesor

    Scuze, postasem intr-adevar o indicatie eronata (asa cum ai observat).

    • 0
  2. user (0)

    Imi faci te rog o rezolvare completa a exercitiului?

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  4. profesor

    Adrianx wrote: Imi faci te rog o rezolvare completa a exercitiului?


    Folosind regulile de calcul cu logaritmi avem:

    a_n = \sum\limits_{k = 1}^n {\log _{\frac{1}{2}} \frac{{k\left( {k + 2} \right)}}{{\left( {k + 1} \right)^2 }}} = \log _{\frac{1}{2}} \prod\limits_{k = 1}^n {\frac{{k\left( {k + 2} \right)}}{{\left( {k + 1} \right)^2 }}} .

    Dar

        \[ \prod\limits_{k = 1}^n {\frac{{k\left( {k + 2} \right)}}{{\left( {k + 1} \right)^2 }}} = \frac{{1 \cdot 3}}{{2^2 }} \cdot \frac{{2 \cdot 4}}{{3^2 }} \cdot \frac{{3 \cdot 5}}{{4^2 }} \cdot \ldots \cdot \frac{{\left( {n - 1} \right)\left( {n + 1} \right)}}{{n^2 }} \cdot \frac{{n\left( {n + 2} \right)}}{{\left( {n + 1} \right)^2 }} \]

    ,

    si dupa simplificari ramane:

        \[ \prod\limits_{k = 1}^n {\frac{{k\left( {k + 2} \right)}}{{\left( {k + 1} \right)^2 }}} = \frac{{\left( {n + 2} \right)}}{{2 \cdot \left( {n + 1} \right)}} \]

    .

    Acum limita se caluleaza cu usurinta:

    {\lim }\limits_{n \to \infty } a_n = \log _{\frac{1}{2}} \left( { {\lim }\limits_{n \to \infty } \frac{{\left( {n + 2} \right)}}{{2 \cdot \left( {n + 1} \right)}}} \right) = \log _{\frac{1}{2}} \frac{1}{2} = 1 \]

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  5. user (0)

    Si limita

        \[ \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{k}{{2^k }}} \]

    cum se rezolva, trebuie calculata suma ? sau trebuie exprimata ca o suma de diferente de fractii pentru a se reduce majoritatea termenilor ?

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