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cosmin_k
Pe: 2013-11-30T17:51:07+02:00 In: In:

limita de siruri

sa se calculeze :

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

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  1. ghioknt
    profesor

    Evident, limita se calculează pentru n->oo; ce rost ar avea condiţia |x|>1, dacă x->oo?
    Fie c_n(x)=1+\frac{1}{x}+\frac{1}{x^{2}}+\frac{1}{x^{3}}+...+\frac{1}{x^{n-1}};\: evident\: c_{n}(x)=\frac{1-\frac{1}{x^n}}{1-\frac{1}{x}}=\frac{x}{x-1}\left ( 1-\frac{1}{x^n} \right ).
    Fie b_n(x)=1+\frac{2}{x}+\frac{3}{x^2}+...+\frac{n-1}{x^{n-2}}+\frac{n}{x^{n-1}};\;\frac{b_n(x)}{x}=\frac{1}{x}+\frac{2}{x^2}+\frac{3}{x^3}+...+\frac{n-1}{x^{n-1}}+\frac{n}{x^{n}}.
    b_n(x)-\frac{b_n(x)}{x}=1+\frac{1}{x}+\frac{1}{x^{2}}+\frac{1}{x^{3}}+...+\frac{1}{x^{n-1}}-\frac{n}{x^{n}}\Rightarrow\,b_{n}(x)\left ( 1-\frac{1}{x} \right )=c_n(x)-\frac{n}{x^n}\Rightarrow b_n(x)=\left ( \frac{x}{x-1} \right )^2\left ( 1-\frac{1}{x^n} \right )-\frac{x}{x-1}\cdot\frac{n}{x^n}.
    Fie a_n(x)=\sum_{k=1}^{n}\frac{k(k+1)}{2x^{k-1}}=\frac{1}{2}\left ( \frac{1\cdot 2}{1}+\frac{2\cdot 3}{x}+\frac{3\cdot 4}{x^2}+...+\frac{(n-1)\cdot n}{x^{n-2}}+\frac{n\cdot (n+1)}{x^{n-1}} \right );\;\frac{a_n(x)}{x}=\frac{1}{2}\left ( \frac{1\cdot 2}{x}+\frac{2\cdot 3}{x^2}+\frac{3\cdot 4}{x^3}+...+\frac{(n-1)\cdot n}{x^{n-1}}+\frac{n\cdot (n+1)}{x^{n}} \right ).
    a_n(x)-\frac{a_n(x)}{x}=\frac{1}{2}\left ( 2+\frac{4}{x}+\frac{6}{x^2}+...\frac{2n}{x^{n-1}}-\frac{n\cdot (n+1)}{x^{n}} \right )\Rightarrow a_n\left ( 1-\frac{1}{x} \right )=b_n-\frac{n\cdot (n+1)}{2x^{n}}\Rightarrow\,a_n=\left ( \frac{x}{x-1} \right )^3\left ( 1-\frac{1}{x^n} \right )-(\frac{x}{x-1})^2\cdot\frac{n}{x^n}-\frac{x}{x-1}\cdot\,\frac{n\cdot (n+1)}{2x^{n}}.
    Dacă |x|>1, atunci \frac{1}{x^n},\;\frac{n}{x^{n}},\;\frac{n^2}{x^n} au limita 0. În concluzie a_n(x) are limita \left ( \frac{x}{x-1} \right )^3 pentru n tinzând la infinit.
    Cu bine, ghioknt.

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