Question

Show the following inequality

1-\dfrac{1}{6n^2}-\dfrac{1}{6n^3} <\displaystyle\sum_{k=1}^n \sqrt[3]{ n^3+k} <1-\dfrac{1}{6n^2}

Collected in the board: Inequality

Steven Zheng posted 1 week ago


Answer

\bigg( 1-\dfrac{k}{3n^3}\bigg) ^3\bigg( 1+\dfrac{k}{n^3}\bigg) <\bigg( 1-\dfrac{k}{3n^3}\bigg) ^3 \bigg( 1+\dfrac{k}{n^3}\bigg) ^3 =\bigg( 1-\dfrac{k^2}{9n^6} \bigg) ^3<1

\bigg( 1-\dfrac{k}{3n^2(n+1)}\bigg) ^3\bigg( 1+\dfrac{k}{n^3}\bigg) >\bigg( 1-\dfrac{k}{n^2(n+1)}\bigg) \bigg( 1+\dfrac{k}{n^3}\bigg) ^3=\dfrac{n^6+n^5+n^2k-k^2}{n^5(n+1)} >1

Taking cube roots of above two equations gives

1-\dfrac{k}{3n^3}<\dfrac{1}{\sqrt[3]{1+\dfrac{k}{n^3}} }< 1-\dfrac{k}{3n^2(n+1)}

Divide all sides by n

\dfrac{1}{n} -\dfrac{k}{3n^4}<\dfrac{1}{\sqrt[3]{n^3+k} }< \dfrac{1}{n} -\dfrac{k}{3n^3(n+1)}

Sum of the inequality where k from 1 to n

1-\dfrac{1}{6n^2}-\dfrac{1}{6n^3} <\displaystyle\sum_{k=1}^n \sqrt[3]{ n^3+k} <1-\dfrac{1}{6n^2}

Steven Zheng posted 1 week ago

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