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C

chonhoi110

đội 3

Có $\dfrac{1}{k^3}<\dfrac{1}{k^3-k}=\dfrac{1}{(k-1)k(k+1)}$

~ >$ A< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{48.49.50}$

Lại có $\dfrac{1}{(n-1)n}-\dfrac{1}{n(n+1)}=\dfrac{2}{(n-1)n(n+1)}$

~ > $A < \dfrac{1}{2}(\dfrac{1}{1.2}-\dfrac{1}{2.3}+...+\dfrac{1}{48.49}-\dfrac{1}{49.50})=\dfrac{1}{4}-\dfrac{1}{2.49.50}<\dfrac{1}{2}$
 
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