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Se considera un sir \( (b_{n})_{n\geq 1} \) un sir de numere pozitive astfel incat \( \lim_{n\to\infty}\frac{b_{n}}{n}=\infty \). Sa se arate ca

a) \( \lim_{n\to\infty}\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{b_{k}}}=0 \).

b) \( \lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{n+b_{k}}=0 \).

a) Aplicam Cesaro Stolz si avem \( \lim_{n\to\infty}\frac{1}{\sqrt{n}}\sum_{k=1}^n {\frac{1}{\sqrt{b_k}}}=\lim_{n\to\infty}\frac{\frac{1}{\sqrt{b_{n+1}}}}{\sqrt{n+1}-\sqrt{n}}=\lim_{n\to\infty}\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{b_{n+1}}}=0 \)


b) \( 0\leq \sum_{k=1}^n {\frac{1}{n+b_k}}\leq\sum_{k=1}^n {\frac{1}{2\sqrt{nb_k}}}\to 0 \) \( (n\to\infty) \) conform punctului a).