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What is the value of $\left\lfloor\sum_{n=1}^{9999} \frac {1} {n^{\frac{1}{4}}}\right\rfloor$ ?

$a. 1332~~b. 1352~~c. 1372~~d.1392$

Attempt: These are some of my ideas :

$1. ~~~~S=\sum_{n=1}^{9999} \dfrac {1} {n^{\frac{1}{4}}} = 1+\dfrac{1}{2^{\frac{1}{4}}}+ \dfrac{1}{3^{\frac{1}{4}}} + \cdots +\dfrac{1}{9999^{\frac{1}{4}}} > 9999 \times \dfrac{1}{10,000^{\frac{1}{4}}}=999.9$

Hence, $S > 999.99$. Not much luck in trying to form a close sandwich from the other side as well.

$2.$I have also tried to form a telescoping series.

$\sum_{n=1}^{9999} \dfrac {1} {n^{\frac{1}{4}}} = \dfrac{n+1-1}{n^{\frac{1}{4}}}= \sum_{n=1}^{9999} \Big \{ \dfrac{n+1}{n^{\frac{1}{4}}} - \dfrac{1}{n^{\frac{1}{4}}}\Big \}$ but this doesn't help much either.

$3.$ Using Integrals. Since $f(n)=\dfrac {1} {n^{\frac{1}{4}}}$ is a monotonically decreasing sequence :

$f(2)+\cdots+f(9999) < \int_1^{9999} \dfrac {dx} {x^{\frac{1}{4}}} < f(1)+\cdots+f(9999)=S$

Can somebody give me some hints please?

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  • $\begingroup$ Well $\frac{1}{n^\frac{1}{4}}$ is equivalent to $n^-\frac{1}{4}$ $\endgroup$
    – user774777
    Commented Apr 21, 2020 at 12:18
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    $\begingroup$ 1332. Replace summation with integral, and 9999~10000.(the error is negligible with this) $\endgroup$
    – user600016
    Commented Apr 21, 2020 at 12:23
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    $\begingroup$ @JosephHulme, does it make any difference? $\endgroup$
    – PinkyWay
    Commented Apr 21, 2020 at 12:23
  • $\begingroup$ @ms._VerkhovtsevaKatya probably not, but it's good to simplify as it can help you visualize it better $\endgroup$
    – user774777
    Commented Apr 21, 2020 at 12:27

3 Answers 3

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You can use the following bounds based on integrals: $$ \frac{4}{3}N^{3/4} - \frac{4}{3} = \int_1^N {\frac{{dx}}{{x^{1/4} }}} \le \sum\limits_{n = 1}^N {\frac{1}{{n^{1/4} }}} \le \int_0^N {\frac{{dx}}{{x^{1/4} }}} = \frac{4}{3}N^{3/4} . $$ In particular, $$ 1331.899 \ldots \le \sum\limits_{n = 1}^{9999} {\frac{1}{{n^{1/4} }}} \le 1333.233 \ldots \, . $$ This is good enough to conlude.

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  • $\begingroup$ Nice explanation. $\endgroup$ Commented Apr 21, 2020 at 12:35
  • $\begingroup$ If you use insted $\int_3^N\frac{dx}{x^{1/4}}+\sum_{n=1}^3\frac{1}{n^{1/4}}\le \sum_{n=1}^N \frac{1}{n^{1/4}}\le \int_2^N\frac{dx}{x^{1/4}}+\sum_{n=1}^3\frac{1}{n^{1/4}}$, you obtain better bounds, sufficient to derive the result without having the fact that it is one of the possibilities $\endgroup$
    – user515010
    Commented Apr 21, 2020 at 12:56
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Cheating:

$$S\lt\int_0^{9999}n^{-1/4}dn=\frac439999^{3/4}=1333.23\cdots$$ and the only compatible choice is $a$.

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Writing this as a sort of Riemann sum, put $$1+\int_2^{10000}\frac{1}{n^{\frac{1}{4}}}dn\quad<\quad\sum_{n=1}^{9999}\frac{1}{n^{\frac{1}{4}}} \quad < \quad 1+\int_1^{9999}\frac{1}{n^{\frac{1}{4}}}dn\\ 1+\frac{4}{3}((\sqrt[4]{10000})^3-(\sqrt[4]{2})^3)\quad <\quad \sum_{n=1}^{9999}\frac{1}{n^{\frac{1}{4}}} \quad <\quad1+\frac{4}{3}((\sqrt[4]{9999})^3-1)\\ 1332.09094\quad <\quad \sum_{n=1}^{9999}\frac{1}{n^{\frac{1}{4}}} \quad <\quad 1332.90000\\ \qquad \qquad\qquad \quad \Biggr\lfloor{\sum_{n=1}^{9999}\frac{1}{n^{\frac{1}{4}}}}\Bigg \rfloor\quad = \quad 1332\qquad\blacksquare $$

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