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FaN.Anduu
Pe: 2019-02-05T21:16:42+02:00 In: In:

limita neterminata

Buna seara,

\lim_{n \to \infty}\dfrac{\dfrac{\ln2}{2}+\dfrac{\ln3}{3}+\cdots+\dfrac{\ln n}{n}}{\ln^2 n}.
Dupa ce am aplicat Stoltz-Cesaro am ajuns la \lim_{n \to \infty}\dfrac{\dfrac{\ln2}{2}+\dfrac{\ln3}{3}+\cdots+\dfrac{\ln n}{n}}{\ln^2 n}=\lim_{n \to \infty}\dfrac{\dfrac{\ln (n+1)}{n+1}}{\ln^2 (n+1)-\ln^2 n}
Limita n-ar trebui sa fie 0? Pentru ca \lim_{n \to \infty}\frac{ln(n+1)}{n+1}=0
Dar nu e 0, e 1/2 si nu stiu cum se ajunge la acest rezultat.

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2 raspunsuri

  1. Felixx
    veteran (III)

    L=\lim_{n\rightarrow \infty }\frac{\frac{ln\left ( n+1 \right )}{n+1}}{ln\frac{n+1}{n}lnn\left ( n+1 \right )}=\lim_{n\rightarrow \infty }\frac{1}{\left ( n+1 \right )ln\frac{n+1}{n}}\lim_{n\rightarrow \infty }\cdot \frac{ln\left ( n+1 \right )}{ln\left ( n+1 \right )+ln\left ( n \right )}=\lim_{n\rightarrow \infty }\frac{1}{ln\left ( \left ( 1+\frac{1}{n} \right )^{n} \right )^{\frac{n+1}{n}}}\cdot \lim_{n\rightarrow \infty }\frac{ln\left ( n+1 \right )}{ln\left ( n+1 \right )\left ( 1+\frac{lnn}{ln\left ( n+1 \right )} \right )}=\frac{1}{lne}\cdot \lim_{n\rightarrow \infty }\frac{1}{1+\frac{\frac{1}{n}}{\frac{1}{n+1}}}=1\cdot \frac{1}{2}=\frac{1}{2}

  2. FaN.Anduu

    Multumesc!

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