TITLE: Finding floor of reciprocal sum
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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
  $\ds{0 < {2 \over 3}\int_{1000}^{\infty}{\braces{x} \over x^{5/3}}\,\dd x <
{2 \over 3}\int_{1000}^{\infty}{\dd x \over x^{5/3}} =
{1 \over 100} = 0.01}$.

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