I need some clues, hints for proving that

$$\lim_{n\to\infty} n\frac{\displaystyle \left\{\frac{n}{\sqrt{1}}\right\}+ \left\{\frac{n}{\sqrt{2}}\right\}+ \left\{\frac{n}{\sqrt{3}}\right\}+\cdots +\left\{\frac{n}{\sqrt{n^2}}\right\}}{\displaystyle \left\{\frac{n}{\sqrt[3]{1}}\right\}+ \left\{\frac{n}{\sqrt[3]{2}}\right\}+ \left\{\frac{n}{\sqrt[3]{3}}\right\}+\cdots +\left\{\frac{n}{\sqrt[3]{n^3}}\right\}}=\frac{\pi^2-12}{6\zeta(3)-9}$$

My first idea was to use Stolz–Cesàro theorem, but this doesn't seem an easy thing to do.


1 Answer 1


It might be easier to write it as $$\frac{2-\zeta(2)}{3/2-\zeta(3)}$$

The numerator is $n^2$ times a Riemann sum for $$\int_{0}^1 \left\{\frac{1}{\sqrt{x}}\right\}dx$$

The denominator is $n^3$ times a Riemann sum for:

$$\int_0^1 \left\{\frac{1}{\sqrt[3]{x}}\right\}dx$$

Since $\int_{0}^1\frac{1}{\sqrt{x}}dx = 2$ and $\int_{0}^1\frac{1}{\sqrt[3]{x}}dx = 3/2$ that seems like the right place to start, since $\int_{0}^1 \{f(x)\}dx$ is $\int_0^1\ f(x) dx$ adjusted subtracting a value that can be expressed as a sum.


$$\begin{align}\int_0^1 \left\lfloor \frac{1}{\sqrt[k]{x}}\right\rfloor dx&=\sum_{n=1}^\infty \int_{\frac{1}{(n+1)^k}}^{\frac{1}{n^k}} ndx\\ &=\sum_{n=1}^\infty n\left(\frac{1}{n^k}-\frac{1}{(n+1)^k}\right) = \zeta(k) \end{align}$$

  • $\begingroup$ Oh, yeahhh. (+1) $\endgroup$ Commented Nov 3, 2014 at 14:14

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