A Floor Over A Floor!

Note that for a certain integer $k$ :

$\displaystyle \lfloor x \rfloor = k \; \;\;\; \text{if} \;\; x\in [k,k+1)$

Let

$\displaystyle k\leq \sqrt[4]{n} < k+1$

Where $k$ is an integer ranging between 1 and $\infty$ .

So for a certain $k$ :

$\displaystyle k^4 \leq n < (k+1)^4$

So our sum can be written as:

$\displaystyle S = \sum_{k=1}^{\infty} \sum_{n=k^4}^{(k+1)^4-1} \frac{\lfloor \sqrt{n} \rfloor }{k^7}$

Now focus on the sum:

$\displaystyle S' = \sum_{n=k^4}^{(k+1)^4-1} \lfloor \sqrt{n} \rfloor$

Let

$\displaystyle m\leq \sqrt{n} < m+1$

Where $m$ is an integer ranging between $k^2$ and $(k+1)^2-1$ . (This to maintain the interval of $n$ in $S'$ .)

So for a certain $m$ :

$\displaystyle m^2 \leq n < (m+1)^2$

So $S'$ can be written as:

$\displaystyle \sum_{m=k^2}^{(k+1)^2 -1 } \sum_{n=m^2}^{(m+1)^2-1 } m$

$\displaystyle = \sum_{m=k^2}^{(k+1)^2 -1 } (m)((m+1)^2-1 - m^2 +1)$

This Sum can be easily manipulated (maybe tedious) using the formulas :

$\displaystyle \sum_{n=1}^{k} n = \frac{k(k+1)}{2}$

$\displaystyle \sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6}$

To get at last the value of $S'$ :

$\displaystyle S' = \frac{5}{3} k + 7k^2 + \frac{34}{3} k^3 + 10k^4 + 4k^5$

Now substitute this in $S$ and Use the fact:

$\displaystyle \zeta(n) = \sum_{k=1}^{\infty} \frac{1}{k^n}$

To get:

$\displaystyle \boxed{S = 4\zeta (2) + 10\zeta (3) + \frac{34}{3} \zeta (4) + 7 \zeta (5) +\frac{5}{3} \zeta (6) }$