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.
 A: 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.
A: 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 $27$.
A: 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, $\displaystyle \left\lfloor \sum_{k=1}^{1000}\frac{1}{\sqrt[3]{k^2}}\right\rfloor=27$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\left\lfloor {1 \over \root[3]{1}} +
{1 \over \root[3]{2^{2}}} + {1 \over \root[3]{3^2}} + \cdots +
{1 \over \root[3]{1000^{2}}}\right\rfloor} =
\left\lfloor{\sum_{k = 1}^{1000}{1 \over
k^{\color{red}{2/3}}}}\right\rfloor
\\[5mm] = &\
\left\lfloor{\zeta\pars{\color{red}{2 \over 3}} - {1000^{1 - \color{red}{2/3}} \over \color{red}{2/3} - 1} +
\color{red}{2 \over 3}\int_{1000}^{\infty}{\braces{x} \over x^{\color{red}{2/3} + 1} }\,\dd x}\right\rfloor.\quad
\pars{~\zeta:\ Riemman\ Zeta\ Function~}
\end{align}
where I used a
Zeta Function Identity.
Then,
\begin{align}
&\bbox[10px,#ffd]{\left\lfloor {1 \over \root[3]{1}} +
{1 \over \root[3]{2^{2}}} + {1 \over \root[3]{3^2}} + \cdots +
{1 \over \root[3]{1000^{2}}}\right\rfloor}
\\[5mm] = &\
\left\lfloor{\zeta\pars{2 \over 3} + 30 +
\color{red}{2 \over 3}\int_{1000}^{\infty}{\braces{x} \over x^{5/3}}
\,\dd x}\right\rfloor
\end{align}

Note that $\ds{\zeta\pars{2/3} \approx -2.4476}$

  and
  $\ds{0 < {2 \over 3}\int_{1000}^{\infty}{\braces{x} \over x^{5/3}}\,\dd x <
{2 \over 3}\int_{1000}^{\infty}{\dd x \over x^{5/3}} =
{1 \over 100} = 0.01}$.

$$
\implies
\bbox[10px,#ffd]{\left\lfloor {1 \over \root[3]{1}} +
{1 \over \root[3]{2^{2}}} + {1 \over \root[3]{3^2}} + \cdots +
{1 \over \root[3]{1000^{2}}}\right\rfloor} = \bbx{27}
$$
