Limita integrala

Post by **Cezar Lupu** » Sat Mar 08, 2008 6:22 pm

Sa se calculze
\( \lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{n+k} \).

Radu Titiu · Post by **Radu Titiu** » Sat Mar 08, 2008 6:36 pm

\( \lim_{n\to\infty}\sum_{k=1}^n \frac{1}{n+k}=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n \frac{1}{1+\frac{k}{n}}=\int_{0}^1 \frac{1}{1+x}dx=\ln(1+1)-\ln(1+0)=\ln2 \)

Virgil Nicula · Post by **Virgil Nicula** » Sat Mar 08, 2008 7:50 pm

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.

Radu Titiu · Post by **Radu Titiu** » Sat Mar 08, 2008 8:22 pm

Se pot vedea si alte solutii aici