floor sum

[man111]

Calculate Sum of [tex]\left[\frac{1}{\sqrt[3]{4}}+\frac{1}{\sqrt[3]{5}}+\frac{1}{\sqrt[3]{6}}+................+\frac{1}{\sqrt[3]{1000000}}\right][/tex]

where [tex]\left[x\right] =[/tex] Greatest Integer function

[martin123456]

Denote [tex]f(x)=\frac{1}{\sqrt[3]{x}}[/tex] and let the domain of [tex]x[/tex] is [tex](0,+\infty)[/tex]. Then [tex]f'(x)=-\frac{1}{3x\sqrt[3]{x}}<0 \Rightarrow f(x) \downarrow[/tex]. Also [tex]f''(x)=\frac{4}{9x^2\sqrt[3]{x}}>0[/tex] so [tex]f(x)[/tex] is convex.

For [tex]x \in [n-1, n][/tex] [tex]f(x)[/tex] is continuous and from the mean value theorem we get that [tex]\exist \xi\in [n-1, n]:\hspace{2mm}\int_{n-1}^{n}{f(x)dx}=f(\xi)(n-n+1)=f(\xi)[/tex]. Since the function is strongly decreasing we get that [tex]f(n)=\frac{1}{\sqrt[3]{n}}<f(\xi)=\int_{n-1}^{n}{f(x)dx}[/tex], e.g. [tex]\frac{1}{\sqrt[3]{n}}<\int_{n-1}^{n}{f(x)dx}[/tex] (1).
From the Jensen's inequlity for convex functions we get [tex]\frac{1}{2}(\frac{1}{\sqrt[3]{n-1}}+\frac{1}{\sqrt[3]{n}})>\frac{1}{\sqrt[3]{n-\frac{1}{2}}}>f(\xi) > \int_{n}^{n+1}{f(x)dx}[/tex] where [tex]\xi \in [n, n+1][/tex] (2).

Further, [tex]\int_{a-1}^{b}{\frac{1}{\sqrt[3]{x}}dx}=\sum_{n=a}^{b}({\int_{n-1}^{n}{f(x)dx}})>\sum_{n=a}^{b}{\frac{1}{\sqrt[3]{n}}[/tex]. Summing (2) for [tex]n \in [a+1,b][/tex] we get [tex]\frac{1}{2}(\frac{1}{\sqrt[3]{a}}+\frac{1}{\sqrt[3]{b-1}})+\sum_{n=a+1}^{b-2}{\frac{1}{\sqrt[3]{n}}}>\int_{a+1}^{b-2}\frac{1}{\sqrt[3]{x}}dx[/tex]. So [tex]\frac{1}{2}(\frac{1}{\sqrt[3]{a-1}}+\frac{1}{\sqrt[3]{b+1}})+\sum_{n=a}^{b}{\frac{1}{\sqrt[3]{n}}}>\int_{a}^{b}\frac{1}{\sqrt[3]{x}}dx[/tex]. And so
[tex]\int_{a-1}^{b}{\frac{1}{\sqrt[3]{x}}dx}>\sum_{n=a}^{b}{\frac{1}{\sqrt[3]{n}}}>\int_{a}^{b}{\frac{1}{\sqrt[3]{x}}dx}-\frac{1}{2}(\frac{1}{\sqrt[3]{a-1}}+\frac{1}{\sqrt[3]{b+1}})[/tex].

Now substitude [tex]a=4[/tex] and [tex]b=1000000[/tex] and just calculate the integrals to get the result.