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) \( \lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{n+b_{k}}=0 \).
Un sir b_n cu limita b_n/n infinita
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Un sir b_n cu limita b_n/n infinita
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
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Marius Mainea
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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).
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).