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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.

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  • $\begingroup$ Note the definition of the Generalized Harmonic numbers: $$H_N^{(s)}=\sum_{n=1}^{N}\frac1{n^s}$$ Your value is $$\lfloor H_{1000}^{(2/3)} \rfloor$$ $\endgroup$ – clathratus Apr 28 at 18:25
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    $\begingroup$ I don’t think there is something telescopic here. Comparing this sum with an integral, and bounding the error term might help. $\endgroup$ – DINEDINE Apr 28 at 18:26
  • $\begingroup$ In particular, you can check the answer of the user hypergeometric at math.stackexchange.com/questions/2570782/…. Follow his idea in your case. You can also work with Abel (partial) summation to estimate your sum. $\endgroup$ – Stelios Sachpazis Apr 28 at 18:28
  • $\begingroup$ The answer is $27$. See here $\endgroup$ – clathratus Apr 28 at 18:30
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    $\begingroup$ Nice when all the different answers get the same result. $\endgroup$ – marty cohen Jun 17 at 20:57
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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, $\displaystyle \left\lfloor \sum_{k=1}^{1000}\frac{1}{\sqrt[3]{k^2}}\right\rfloor=27$.

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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.

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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 $27$.

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

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

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