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Pe: 2014-04-06T12:44:45+03:00 In: In:

limita cu progresie

\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{a_{k} a_{k+1}}=?,
unde (a_k),k∈N\{0}, formeaza o progresie aritmetica cu ratia r≠0.

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  1. PhantomR
    expert (VI)

    Fie s_n=\sum_{k=1}^n\frac{1}{a_ka_{k+1}}=\sum_{k=1}^n\frac{1}{r}\cdot \frac{r}{a_k(a_k+r)}=\frac{1}{r}\sum_{k=1}^n\frac{a_k+r-a_k}{a_k(a_k+r)}=\frac{1}{r}\sum_{k=1}^n\frac{1}{a_k}-\frac{1}{a_k+r}=\frac{1}{r}\sum_{k=1}^n\frac{1}{a_k}-\frac{1}{a_{k+1}}=\frac{1}{r}\left(\frac{1}{a_1}-\frac{1}{a_2}+\frac{1}{a_2}-\frac{1}{a_3}+...+\frac{1}{a_{n-1}}-\frac{1}{a_n}+\frac{1}{a_n}-\frac{1}{a_{n+1}}\right) = \frac{1}{r}\left(\frac{1}{a_1}-\frac{1}{a_{n+1}}\right)=\frac{1}{r}\left( \frac{1}{a_1}-\frac{1}{a_1+(n-1)r}\right). De aici, \lim_{n\to\infty} s_n= \frac{1}{r}\left( \frac{1}{a_1}-0\right)=\frac{1}{r}\cdot \frac{1}{a_1}=\frac{1}{ra_1}.

    Deci 

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    .

    NOTA: Cand aveti sume ce au termenii de forma \frac{1}{x(x+k)}, s-ar putea sa va foloseasca urmatorul artificiu: \frac{1}{x(x+k)}=\frac{1}{k}\cdot\frac{k}{x(x+k)}=\frac{1}{k}\cdot \frac{x+k-x}{x(x+k)}=\frac{1}{k}\left( \frac{x+k}{x(x+k}}-\frac{x}{x(x+k)}\right)=\frac{1}{k}\left( \frac{1}{x}-\frac{1}{x+k}\right).

    Era mai rapid sa scriem \frac{1}{a_ka_{k+1}}=\frac{1}{r}\cdot\frac{r}{a_ka_{k+1}}=\frac{1}{r}\cdot\frac{a_{k+1}-a_k}{a_ka_{k+1}}=\frac{1}{r}\left( \frac{1}{a_k}-\frac{1}{a_{k+1}}\right), dar poate se vede mai bine mersul in forma initiala.

    NOTA: Avem \lim_{n\to\infty} \frac{1}{a_1+(n-1)r}=0, caci daca r>0, numitorul tinde la +\infty, iar daca r<0, el are limita -\infty. In ambele cazuri, fractia are limita 0.

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