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

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?

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

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.
• 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
$$S\lt\int_0^{9999}n^{-1/4}dn=\frac439999^{3/4}=1333.23\cdots$$ and the only compatible choice is $$a$$.
Writing this as a sort of Riemann sum, put $$1+\int_2^{10000}\frac{1}{n^{\frac{1}{4}}}dn\quad<\quad\sum_{n=1}^{9999}\frac{1}{n^{\frac{1}{4}}} \quad < \quad 1+\int_1^{9999}\frac{1}{n^{\frac{1}{4}}}dn\\ 1+\frac{4}{3}((\sqrt[4]{10000})^3-(\sqrt[4]{2})^3)\quad <\quad \sum_{n=1}^{9999}\frac{1}{n^{\frac{1}{4}}} \quad <\quad1+\frac{4}{3}((\sqrt[4]{9999})^3-1)\\ 1332.09094\quad <\quad \sum_{n=1}^{9999}\frac{1}{n^{\frac{1}{4}}} \quad <\quad 1332.90000\\ \qquad \qquad\qquad \quad \Biggr\lfloor{\sum_{n=1}^{9999}\frac{1}{n^{\frac{1}{4}}}}\Bigg \rfloor\quad = \quad 1332\qquad\blacksquare$$