Aratati ca daca seria \( \sum_{n=1}^{\infty}a_{n} \) este convergenta, atunci si seria \( \sum_{n=1}^{\infty}a_{n}^{\frac{n}{n+1}} \) este convergenta.
Putnam 1988
Serie convergenta cu exponentul la limita 1
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Cezar Lupu
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Serie convergenta cu exponentul la limita 1
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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Octav Ganea
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Sol 1: Din medii avem \( \sqrt[n+1]{a_n^n} \leq \frac{(n-1)\cdot a_n + 2\cdot \sqrt{a_n}}{n+1} \leq a_n+a_n + \frac{1}{(n+1)^2} \).
Sol 2: Daca \( a_n \geq \frac {1}{2^{n+1}} \) atunci \( \sqrt[n+1]{a_n^n} \leq \frac{1}{2^n} \) , daca nu atunci \( \frac{a_n}{\sqrt[n]{a_n}} \leq 2\cdot a_n \); astfel seria mare se imparte in doua subserii convergente fiecare.
Sol 2: Daca \( a_n \geq \frac {1}{2^{n+1}} \) atunci \( \sqrt[n+1]{a_n^n} \leq \frac{1}{2^n} \) , daca nu atunci \( \frac{a_n}{\sqrt[n]{a_n}} \leq 2\cdot a_n \); astfel seria mare se imparte in doua subserii convergente fiecare.