Finding floor of reciprocal sum

One approach is to approximate the sum with an integral, and show that the error is bounded by a number less than 1.

In particular, let $$H_{1000}^{(2/3)} = \sum_{k=1}^{1000} k^{-2/3}, \quad I = \int_{x=1}^{1000} x^{-2/3} \, dx.$$ Then we know $$I \le H_{1000}^{(2/3)} < I+1.$$ But $I = 27$, and we are done.

You can bound this summation by two integrals; $$\int_1^{1001} x^{-2/3}\mathrm{d}x\lt\sum_{k=1}^{1000}k^{-2/3}\lt1+\int_1^{1000}x^{-2/3}\mathrm{d}x$$ Hence we have $$27\lt3\sqrt[3]{1001}-3\lt\sum_{k=1}^{1000}k^{-2/3}\lt28$$ So as the value of the sum is strictly between $27$ and $28$, the floor of the sum is $27$.

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, $\displaystyle \left\lfloor \sum_{k=1}^{1000}\frac{1}{\sqrt[3]{k^2}}\right\rfloor=27$.