Find the greatest integer less than or equal to $\sum\limits_{n=1}^{9999} \frac {1} {n^{\frac 1 4}}.$

Question

Find the greatest integer less than or equal to $$\sum\limits_{n=1}^{9999} \frac {1} {n^{\frac 1 4}}.$$

I have tried but failed. Is there any elementary way to find this out? Any help will be highly appreciated.

Thank you very much for your valuable time.

Answer 1

Since $x^{-1/4}$ is decreasing, then

$$\underbrace{\int_1^{10,000}\frac1{x^{1/4}}\,dx}_{1,332} \le \sum_{n=1}^{9999}\frac1{n^{1/4}}\le 1+\underbrace{\int_1^{9999}\frac1{x^{1/4}}\,dx}_{=\frac43 ((9999)^{3/4}-1)}$$

Can you convince yourself that the right-hand side is less than 1,333?

Answer 2

Consider the function $f(x)=\frac{4}{3} x^{\frac{3}{4}}, f:[k, k+1] \rightarrow \mathbb{R}$, and we apply the MVT theoreme, so we can write: $\frac{1}{\sqrt[4]{k+1}}<f(k+1)-f(k)<\frac{1}{\sqrt[4]{k}}$. Denote $S_{n}=\sum_{k=1}^{n} \frac{1}{\sqrt[4]{k}}$, so after the summarization we get: $S_{n}-1<f(n)-\int(1)<S_{n}-\frac{1}{\sqrt[4]{n}} \Leftrightarrow \frac{4}{3} n^{\frac{3}{4}}+\frac{1}{\sqrt[4]{n}}-\frac{4}{3}<S_{n}<\frac{4}{3} n^{\frac{3}{2}}-\frac{1}{3}$, and replace $n=9999$.