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Find the integral part of $\sum_{n=1}^{10^9} {n^{- \frac {2}{3}}}$.

SOURCE : Inequalites - CRUX (Page Number 3 ; Question Number 160)

Wolfram Alpha gives the answer as $\approx 2997.55$ .

For upper bound as $k$ instead of $10^9$, it gives a general formula as $H_k^{ \big({\frac {2}{3}\big) }}$, where $H_n^{(r)}$ i the generalized harmonic number.

So, the answer is essentially $H_{10^9} ^{ \big({\frac {2}{3}\big) }}$.

Can anyone explain how to obtain this answer?

Any help would be gratefully acknowledged :).

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    It is the definition ... – 2017-01-27
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    @SimplyBeautifulArt What do you mean ? – 2017-01-27
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    Define generalized harmonic number – 2017-01-27
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    Say, what does "integral part" mean here? – 2017-01-27

2 Answers 2

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By Euler-Maclaurin formula:

$$\sum_{n=1}^{10^9}\frac1{n^{2/3}}\approx\zeta(2/3)+\int_0^{10^9}\frac1{x^{2/3}}\ dx+\frac1{2(10^9)^{2/3}}\\=\zeta(2/3)+3(10^9)^{1/3}+\frac12(10^{-6})\approx2997.552$$

As for the generalized harmonic number, they are equal by definition, which is rather useless except for the fact that you can now Google about the generalized harmonic number.

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    :-/ Why the downvote? I cannot help if I did not understand what "integral part" meant, and I even asked for clarification with no response after 3 hours. – 2017-01-27
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    The downvote is not from me.... I just upvoted you..... Nice answer.. :) – 2017-01-28
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We have, by integral-series estimation, $$\int_1^{10^9+1} x^{-2/3} \,dx < \sum_{1\leq k\leq 10^9} k^{-2/3}<1^{-2/3}+\int_1^{10^9} x^{-2/3} \,dx$$ So $$\left.3\sqrt[3]{x}\right|^{10^9+1}_1 < \sum_{1\leq k\leq 10^9} k^{-2/3} < 1+\left.3\sqrt[3]{x}\right|^{10^9}_1$$ So $$2997<3\sqrt[3]{10^9 + 1} - 3 < \sum_{1\leq k\leq 10^9} k^{-2/3} < 2998$$ So the integer part of the sum is $\boxed{2997}$

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    Oh... integer part! Jeez I am stupid... – 2017-01-27