O limita cu sirul armonic
Posted: Wed Sep 26, 2007 9:02 pm
Sa se calculeze \( \lim_{n\to\infty} e^{H_{n+1}}-e^{H_{n}} \), unde \( H_{n}=1+\frac{1}{2}+\ldots +\frac{1}{n} \) este sirul armonic.
E echivalenta cu :
\( \lim_{n}\frac{e^{H_{n}}}{n+1}\cdot \frac{e^{\frac{1}{n+1}}-1}{\frac{1}{n+1}}= \lim_{n} (e^{H_{n}-\ln(n)})\cdot \frac{n}{n+1}=e^c \) ,unde c este constanta lui Euller