[SOLVED] Series Inequality

[anemone]

Prove that $\sqrt[3]{\dfrac{2}{1}}+\sqrt[3]{\dfrac{3}{2}}+\cdots+\sqrt[3]{\dfrac{996}{995}}-\dfrac{1989}{2}<\dfrac{1}{3}+\dfrac{1}{6}+\cdots+\dfrac{1}{8961}$.

 

[anemone]

Spoiler: Solution of other

Note that

$\dfrac{3k+1}{3k}+\dfrac{3k+2}{3k+1}+\dfrac{3k+3}{3k+2}=3+\dfrac{1}{3k}+\dfrac{1}{3k+1}+\dfrac{1}{3k+2}$ and

$\dfrac{3k+1}{3k}\cdot\dfrac{3k+2}{3k+1}\cdot\dfrac{3k+3}{3k+2}=\dfrac{k+1}{k}$,

by the Arithmetic-Geometric Mean inequality, we have

$3+\dfrac{1}{3k}+\dfrac{1}{3k+1}+\dfrac{1}{3k+2}>3\sqrt[3]{\dfrac{k+1}{k}}$

Then

$\displaystyle 3\sum_{k=1}^{995} \sqrt[3]{\dfrac{k+1}{k}}<3\sum_{k=1}^{995}\left( 3+\dfrac{1}{3k}+\dfrac{1}{3k+1}+\dfrac{1}{3k+2}\right)$

and hence

$\displaystyle\begin{align*} 3\sum_{k=1}^{995} \sqrt[3]{\dfrac{k+1}{k}}-\dfrac{1989}{2}&<995-\dfrac{1989}{2}+\sum_{k=1}^{995}\dfrac{1}{3}\left( \dfrac{1}{3k}+\dfrac{1}{3k+1}+\dfrac{1}{3k+2}\right)\\&=\dfrac{1}{2}+\left(\dfrac{1}{9}+\dfrac{1}{12}+\cdots+\dfrac{1}{8961}\right)\\&=\dfrac{1}{3}+\dfrac{1}{6}+\left(\dfrac{1}{9}+\dfrac{1}{12}+\cdots+\dfrac{1}{8961}\right)\end{align*}$