Let the sequence $a_n$, $n\geq2$, $a_n=\frac{\sqrt[3]{n^3+n^2-n-1}}{n} $. Find the greatest natural number $k$ ,such that
$a_2 \cdot a_3 \cdot . . .\cdot a_k <8$
Note that $a_{n}=\frac{(n-1)^{1 / 3}(n+1)^{2 / 3}}{n}$
So $a_{2} a_{3} \cdots a_{k}=\frac{((k-1) !)^{1 / 3}\left(\frac{(k+1) !}{2}\right)^{2 / 3}}{k !}=\frac{(k-1) !\left(\frac{k^{2}+k}{2}\right)^{2 / 3}}{k !}$
This simplifies to
$
\left(\frac{k^{2}+k}{2 k^{3 / 2}}\right)^{2 / 3}=\left(\frac{\sqrt{k}}{2}+\frac{1}{2 \sqrt{k}}\right)^{2 / 3}
$
$\mathrm{Now}\left(\frac{\sqrt{k}}{2}+\frac{1}{2 \sqrt{k}}\right)^{2 / 3}<8 \quad$ i.e. $ \sqrt{k}+\frac{1}{\sqrt{k}}<32 \sqrt{2}$
For $k = 2039$ the inequality holds whereas when $k= 2040$ the inequality is dissatisfied. As $\sqrt{k} + \frac{1}{\sqrt{k}}$ is a increasing function on $k \in N$, the required value is
Let $f(k) = \frac{\sqrt{k}}{2}+\frac{1}{2 \sqrt{k}}$
We prove that if $x,y \in N $ with $x>y$, $f(x) > f(y)$
Note that $f(x) - f(y) = (\sqrt{x}-\sqrt{y})(1-\frac{1}{\sqrt{xy}})$.As $x > y> 1$ we get that the RHS is positive
This proving the required condition
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