<s>[INST] Finding floor of reciprocal sum 
Evaluation of
$$\bigg \lfloor \frac{1}{\sqrt[3]{1}}+\frac{1}{\sqrt[3]{2^2}}+\frac{1}{\sqrt[3]{3^2}}+\cdots +\frac{1}{\sqrt[3]{(1000)^2}}\bigg\rfloor$$
Where $\lfloor x\rfloor $ is the floor of $x$

Try: It seems like we can solve it using Telescopic sums and that the sum lies between $2$ Telescopic sums, but could not figure out how to solve it.
Could someone help me to solve it? Thanks.
 [/INST] Note that $\displaystyle \sum_{k=1}^{1000}\frac{3}{\sqrt[3]{(k+1)^2}+\sqrt[3]{k(k+1)}+\sqrt[3]{k^2}}<\sum_{k=1}^{1000}\frac{1}{\sqrt[3]{k^2}}<1+\sum_{k=2}^{1000}\frac{3}{\sqrt[3]{(k-1)^2}+\sqrt[3]{k(k-1)}+\sqrt[3]{k^2}}$.
$\displaystyle \sum_{k=1}^{1000}\frac{3}{\sqrt[3]{(k+1)^2}+\sqrt[3]{k(k+1)}+\sqrt[3]{k^2}}=\sum_{k=1}^{1000}\frac{3(\sqrt[3]{k+1}-\sqrt[3]{k})}{(k+1)-k}=3(\sqrt[3]{1001}-1)>27$
$\displaystyle 1+\sum_{k=2}^{1000}\frac{3}{\sqrt[3]{(k-1)^2}+\sqrt[3]{k(k-1)}+\sqrt[3]{k^2}}=1+\sum_{k=2}^{1000}\frac{3(\sqrt[3]{k}-\sqrt[3]{k-1})}{(k-1)-k}=1+3(\sqrt[3]{1000}-1)=28$
So, $\displaystyle \left\lfloor \sum_{k=1}^{1000}\frac{1}{\sqrt[3]{k^2}}\right\rfloor=27$.
</s>