What is the value of $$[(1^ {-2/3}) + (2^ {-2/3}) + (3^ {-2/3}) + … + (1000^ {-2/3})]?$$ where $[x]$ stands for the greatest integer function.
P. S : I ran a code on my PC to find that the quantity inside the [ ] function is approximately equal to $27.5$ , thus $[27.5] = 27$. I am however Looking for a mathematical way to approximate the sum, instead of having run a code.
On
Making the problem more general $$S_n=\sum_{k=1}^n k^{-\frac 23}=H_n^{\left(\frac{2}{3}\right)}$$ where appears generalized harmonic numbers.
Using asymptotics $$S_n=3n^{\frac 13}+\zeta \left(\frac{2}{3}\right)+\frac {1 }{2n^{\frac 23} }-\frac {1 }{18n^{\frac 53} }+\frac {1 }{243n^{\frac {11}3} }+\cdots$$ If $n=1000$, $\frac {1 }{2n^{\frac 23} }=\frac 1{200}$ and we can admit that it would not impact the result. So the two first terms are sufficient and $\zeta \left(\frac{2}{3}\right)=-2.44758\approx -\frac 52$ making $$S_{1000}\sim 30 -\frac 52=27.5$$ which is what you obtained.
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Let $$S:=\sum_{k=1}^{1000} k^{-2/3}\ .$$ Then $$S<1+\int_1^{1000}x^{-2/3}\>dx=1+3x^{1/3}\biggr|_1^{1000}=28\ .$$ In the same way $$S>1+\int_2^{1001}x^{-2/3}\>dx=28+\int_{1000}^{1001}x^{-2/3}\>dx-\int_1^2x^{-2/3}\>dx>27\ ,$$ since the first "correcting integral" on the right is smaller than the second, and the second smaller than $1$. It follows that $$\lfloor S\rfloor=27\ .$$
Hint. Note that $$S-1=\sum_{k=1}^{999}(k+1)^{-2/3}< \sum_{k=1}^{999}\int_{k}^{k+1}x^{-2/3}\,dx< \sum_{k=1}^{999}k^{-2/3}=S-\frac{1}{100}$$ where $S=\sum_{k=1}^{1000}k^{-2/3}$.