Summing Up The Nearest!

First note that for a certain integer $k$ :

$\displaystyle \lfloor x \rceil = \left\{ \begin{array}{lr} k & \text{if} \; x \in [k,k+0.5) \\k+1 & \text{if} \; x \in (k+0.5, k+1 ] \end{array} \right.$

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

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

Let us split the interval into two :

$\displaystyle k\leq \sqrt[4]{n} < k+0.5 \; \; \text{and} \; \; k+0.5 < \sqrt[4]{n} <k+1$

Note that we restricted the value at $k+0.5$ , since $\sqrt[4]{n}$ is either irrational or integral. Now rewrite the intervals as:

$\displaystyle k^4\leq n < (k+0.5)^4 \; \; \text{and} \; \; (k+0.5 )^4< n <(k+1 )^4$

$n$ is an integer , so rewrite as:

$\displaystyle k^4\leq n \leq \lfloor (k+0.5)^4 \rfloor \; \; \text{and} \; \; \lceil (k+0.5 )^4 \rceil \leq n <(k+1 )^4$

Therefore our Sum can be written as :

$\displaystyle S= \sum_{n=1}^{\infty} \frac{1}{ \lfloor \sqrt[4]{n} \rceil ^7 }$

$\displaystyle = \sum_{k=1}^{\infty} \bigg( \sum_{n=k^4}^{ \lfloor (k+0.5)^4 \rfloor} \frac{1}{k^7} + \sum_{n=\lceil (k+0.5 )^4 \rceil }^{(k+1)^4 -1} \frac{1}{(k+1)^7} \bigg)$

The work inside the brackets is easy (as we have no $n$ ):

$\displaystyle S= \sum_{k=1}^{\infty} \bigg( \frac{ \lfloor (k+0.5)^4 \rfloor - k^4 +1}{k^7} + \frac{(k+1)^4 -1 - \lceil (k+0.5 )^4 \rceil +1 }{(k+1)^7} \bigg)$

Note that we should find an expression for $\lfloor (k+0.5)^4 \rfloor$ :

$\displaystyle (k+0.5)^4 = k^4 + 2k^3 + \frac{3}{2} k^2 + \frac{1}{2} k +\frac{1}{16}$

$\displaystyle = k^4 + 2k^3 + \frac{k(3k+1)}{2} + \frac{1}{16}$

For any integral $k$ , either $k$ or $3k+1$ will be even, hence :

$\displaystyle \frac{k(3k+1)}{2} \;\;\;\;\text{is an integer}$

And so :

$\displaystyle \lfloor (k+0.5)^4 \rfloor = (k+0.5)^4 - \frac{1}{16}$

Consequently:

$\displaystyle \lceil (k+0.5 )^4 \rceil = (k+0.5)^4 - \frac{1}{16} +1$

Now back to our sum:

$\displaystyle S = \sum_{k=1}^{\infty} \bigg( \frac{ (k+0.5)^4 - \frac{1}{16} - k^4 +1}{k^7} + \frac{(k+1)^4 -1 - (k+0.5)^4 + \frac{1}{16} }{(k+1)^7} \bigg)$

$\displaystyle = \sum_{k=1}^{\infty} \frac{ (k+0.5)^4 - \frac{1}{16} - k^4 +1}{k^7} + \sum_{k=1}^{\infty} \frac{(k+1)^4 -1 - (k+0.5)^4 + \frac{1}{16} }{(k+1)^7}$

Now shift the index of the second series by one:

$\displaystyle = \sum_{k=1}^{\infty} \frac{ (k+0.5)^4 - \frac{1}{16} - k^4 +1}{k^7} + \sum_{k=2}^{\infty} \frac{k^4 -1 - (k-0.5)^4 + \frac{1}{16} }{k^7}$

Note that the second series is zero at $k=1$ :

$\displaystyle = \sum_{k=1}^{\infty} \frac{ (k+0.5)^4 - \frac{1}{16} - k^4 +1}{k^7} + \sum_{k=1}^{\infty} \frac{k^4 -1 - (k-0.5)^4 + \frac{1}{16} }{k^7}$

$\displaystyle = \sum_{k=1}^{\infty} \bigg( \frac{ (k+0.5)^4 - \frac{1}{16} - k^4 +1}{k^7} + \frac{k^4 -1 - (k-0.5)^4 + \frac{1}{16} }{k^7} \bigg)$

$\displaystyle = \sum_{k=1}^{\infty} \frac{ (k+0.5)^4 - (k-0.5)^4}{k^7}$

$\displaystyle = \sum_{k=1}^{\infty} \frac{ 4k^3 +k}{k^7}$

$\displaystyle = 4\sum_{k=1}^{\infty} \frac{1}{k^4} + \sum_{k=1}^{\infty} \frac{1}{k^6}$

$\displaystyle = 4\zeta (4) + \zeta (6)$

$\displaystyle \boxed{ S = \frac{ 2}{45} \pi^4 + \frac{1}{945}\pi^6 }$

I used the general expression of nint function.

$\displaystyle S_N(s) = \sum_{n=1}^{\infty} \frac{1}{(\lfloor n^{1/N} \rceil)^s}$

Expressions are available for $s>3$

I also used . $S_4(s) = 4\zeta(s-3) + \zeta(s-1)$

For s = 7 , we have = $\boxed{\frac{2\pi^4}{45} + \frac{\pi^6}{945}}$