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Felicia_b
Pe: 2013-07-10T16:20:50+03:00 In: In:

Limita cu e

Fie doua siruri (a_n) si (b_n) astfel

    \[{a_n} = \sum\limits_{k = 1}^n {\frac{1}{{\sqrt {{n^2} + k} }}} \;\;;{b_n} = n\]

.Sa se calculeze

    \[{\lim }\limits_{n \to \infty } {\left( {{a_n}} \right)^{{b_n}}}\]

Similare

5 raspunsuri

  1. xor_NTG
    maestru (V)

    Eu asa as face:

        \[ e^{n\ln \left( {\sum\limits_{k = 1}^n {\frac{1}{{\sqrt {n^2 + k} }}} } \right)} = e^{n\ln \left( {\frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{\sqrt {1 + \frac{k}{{n^2 }}} }}} } \right)} = e^{n\left[ {\ln \left( {\frac{1}{n}} \right) + \ln \left( {\sum\limits_{k = 1}^n {\frac{1}{{\sqrt {1 + \frac{k}{{n^2 }}} }}} } \right)} \right]} = e^{n\left( {\ln 1 - \ln n} \right) + n\ln n} = e^{n\ln n - n\ln n} = e^0 = 1 \]

  2. George_Gaumont
  4. TheodorMunteanu

    xor_NTG wrote: Eu asa as face:

        \[ e^{n\ln \left( {\sum\limits_{k = 1}^n {\frac{1}{{\sqrt {n^2 + k} }}} } \right)} = e^{n\ln \left( {\frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{\sqrt {1 + \frac{k}{{n^2 }}} }}} } \right)} = e^{n\left[ {\ln \left( {\frac{1}{n}} \right) + \ln \left( {\sum\limits_{k = 1}^n {\frac{1}{{\sqrt {1 + \frac{k}{{n^2 }}} }}} } \right)} \right]} = e^{n\left( {\ln 1 - \ln n} \right) + n\ln n} = e^{n\ln n - n\ln n} = e^0 = 1 \]


    poate ne explicati cum ati ajuns la a 3-a egalitate

  5. xor_NTG
    maestru (V)

    Da. E mult mai complicat decat pare…

  6. DD
    profesor

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