O limita cu sirul armonic
Posted: Sun Sep 30, 2007 11:58 am
Sa se calculeze \( \lim_{n\to\infty}(\sqrt[n] {n})^{H_{n}} \), unde \( H_{n}=1+\frac{1}{2}+\ldots+\frac{1}{n} \).
\( L=\lim_{n} \left( 1+\sqrt[n]{n}-1 \right)^{\frac{1}{\sqrt[n]{n}-1}\cdot (\sqrt[n]{n}-1)H_{n}}=e^{\lim_{n} (\sqrt[n]{n}-1)H_{n}} \)
\( \lim_{n} (\sqrt[n]{n}-1)H_{n}= \lim_{n}\frac{e^{\frac{ln(n)}{n}}-1}{\frac{ln(n)}{n}}\cdot \frac{ln(n)^2}{n}\frac{H_{n}}{ln(n)}=\lim_{n}\frac{ln(n)^2}{n}=0 \)
deci L=1