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Andreea09
user (0)
Pe: 2012-12-02T08:23:01+02:00 In: In:

convergenta

Am intampinat niste dificultati in rezolvarea acestor 2 exercitii. Daca va puteti uita, voi fi recunoscatoare:

Sa se studieze convergenta sirurilor:

    \[\begin{array}{l} a){a_n} = (1 - \frac{1}{{{2^2}}})(1 - \frac{1}{{{3^2}}})...(1 - \frac{1}{{{{(n + 1)}^2}}});\\ b){a_n} = \frac{1}{2}\frac{3}{4}\frac{5}{6}....\frac{{2n - 1}}{{2n}}; \end{array}\]

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4 raspunsuri

  1. sandy_sc
    maestru (V)

    a)
    Observi ca fiecare paranteza reprezinta o diferenta de patrate.O descompui Continuarea in atasament

    • 0
  2. PhantomR
    expert (VI)

    Pentru b)

    • 0
  4. ali
    maestru (V)

    b) mărginire
    \begin{array}{l} {b_n} = \frac{{\prod {2k - 1} }}{{\prod {2k} }}\\ 1 \times 3 < {2^2}\\ 3 \times 5 < {4^2}\\ 5 \times 7 < {6^2}\\ ........\\ \left( {2n - 3} \right)\left( {2n - 1} \right) < {\left( {2n - 2} \right)^2}\\ \left( {2n - 1} \right)\left( {2n + 1} \right) < {\left( {2n} \right)^2}\\ \Rightarrow \frac{{{{\left( {\prod {2k - 1} } \right)}^2} \times \left( {2n + 1} \right)}}{{{{\left( {\prod {2k} } \right)}^2}}} < 1 \Rightarrow \frac{{\prod {2k - 1} }}{{\prod {2k} }} < \frac{1}{{\sqrt {2n + 1} }} < 1,\,n \in {N^*} \end{array}

    • 0
  5. PhantomR
    expert (VI)

    Multumesc mult pentru postare, domnule ali.

    • 0
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