# 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.

• Note the definition of the Generalized Harmonic numbers: $$H_N^{(s)}=\sum_{n=1}^{N}\frac1{n^s}$$ Your value is $$\lfloor H_{1000}^{(2/3)} \rfloor$$ – clathratus Apr 28 at 18:25
• I don’t think there is something telescopic here. Comparing this sum with an integral, and bounding the error term might help. – HAMIDINE SOUMARE Apr 28 at 18:26
• 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. – Stelios Sachpazis Apr 28 at 18:28
• The answer is $27$. See here – clathratus Apr 28 at 18:30
• Nice when all the different answers get the same result. – marty cohen Jun 17 at 20:57

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

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 $$27$$.


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

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