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Limita integrala
Posted: Sat Mar 08, 2008 6:22 pm
by Cezar Lupu
Sa se calculze
\( \lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{n+k} \).
Posted: Sat Mar 08, 2008 6:36 pm
by Radu Titiu
\( \lim_{n\to\infty}\sum_{k=1}^n \frac{1}{n+k}=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n \frac{1}{1+\frac{k}{n}}=\int_{0}^1 \frac{1}{1+x}dx=\ln(1+1)-\ln(1+0)=\ln2 \)
Posted: Sat Mar 08, 2008 7:50 pm
by Virgil Nicula
Exista o solutie simpla la nivelul clasei a XI - a (consecinta a numarului \( e \) ).
Moderatorii pot posta aceasta problema si la clasa a XI - a.
Posted: Sat Mar 08, 2008 8:22 pm
by Radu Titiu
Se pot vedea si alte solutii
aici