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Alex Stoica
Pe: 2018-07-04T13:36:05+03:00 In: In:

UTCN 371

As avea nevoie de putin ajutor la exercitiul 371.

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  1. PhantomR
    expert (VI)

    Limita de calculat arata cam urat. Sa notam b_n=\ln \sqrt[n^2]{a_n} = \frac{\ln a_n}{n^2}. Conform teoremei Cesaro-Stolz, avem \lim b_n = \frac{\ln a_{n+1} - \ln a_n}{(n+1)^2-n^2} = \frac{\ln a_{n+1} - a_n}{2n+1}.

    Sa vedem acum cum putem trece la logaritmi in relatia initiala. Inductiv, a_n>0,\forall n. Logaritmam si notam y_n=\ln a_n spre a obtine
    y_{n+2}+y_n = \ln (1+\frac{1}{n})^n + 2y_{n+1} \ (1).
    Cu notatia noua, limita de calculat este acum \lim \frac{y_{n+1} - y_n}{2n+1}. Sa observam ca putem rearanja termenii in (1) astfel:
    y_{n+2}-y_{n+1} - (y_{n+1} - y_n) = \ln (1+\frac{1}{n})^n. Notand acum z_n = y_{n+1}-y_n, limita de calculat devine \lim \frac{z_n}{2n+1}, iar relatia anterioara se rescrie ca z_{n+1} - z_n = \ln (1+\frac{1}{n})^n.

    Ei bine, forma celor doua relatii la care am ajuns sugereaza sa mai aplicam inca o data Cesaro-Stolz spre a obtine \lim b_n = \lim \frac{z_{n+1} - z_n}{2(n+1) + 1 - (2n+1)}= \lim \frac{\ln (1+\frac{1}{n})^n}{2}= \frac{1}{2}.

    In final, \lim a_n = e^{\lim b_n} = \sqrt{e}.

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