Is it true that $\left\lfloor\sum_{s=1}^n\operatorname{Li}_s\left(\frac 1k \right)\right\rfloor\stackrel{?}{=}\left\lfloor\frac nk \right\rfloor$ While studying polylogarithms I observed the following.
Let $n>0$ and $k>1$ be integers. Is the following statement true?
$$\left\lfloor \sum_{s=1}^n \operatorname{Li}_s\left( \frac{1}{k} \right) \right\rfloor \stackrel{?}{=} \left\lfloor \frac{n}{k} \right\rfloor $$
If it is, then how could we prove it? If not, give a counterexample.
 A: We use the series expansion for the polylogarithm. 
Note that 
$$\sum_{s=1}^{n}\operatorname{Li_{s}}\left(\frac{1}{k}\right) = 
\sum_{s=1}^{n}\sum_{j=1}^{\infty}\frac{1}{j^sk^j} = \sum_{j=1}^{\infty}\sum_{s=1}^{n}\frac{1}{j^{s}k^j}$$
we are able to interchange the summations because each term is positive, which implies that the sum is absolutely convergent.
Now separating out the first term gives
$$\frac{n}{k} + \sum_{j=2}^{\infty}\sum_{s=1}^{n}\frac{1}{j^sk^j}$$
We may then use the formula for a geometric sum to bring this to
$$\frac{n}{k} + \sum_{j=2}^{\infty}\frac{1}{j}\frac{1 - \frac{1}{j^{n+1}}}{\left(1 - \frac{1}{j}\right)k^j} = \frac{n}{k} + \sum_{j=2}^{\infty}\frac{j^{n+1} - 1}{j^{n+1}\left(j-1\right)k^j} = $$
$$\frac{n}{k} + \sum_{j=2}^{\infty}\frac{1}{\left(j-1\right)k^j} - \sum_{j=2}^{\infty}\frac{1}{j^{n+1}\left(j-1\right)k^j}$$
Now since clearly 
$$\sum_{j=2}^{\infty}\frac{1}{\left(j-1\right)k^{j}} > \sum_{j=2}^{\infty}\frac{1}{j^{n+1}\left(j-1\right)k^j}$$
we have
$$\frac{n}{k} < \sum_{s=1}^{n}\operatorname{Li}_{s}\left(\frac{1}{k}\right) <
\frac{n}{k} + \sum_{j=2}^{\infty}\frac{1}{\left(j-1\right)k^j} = 
\frac{n}{k} + \frac{1}{k}\sum_{j=1}^{\infty}\frac{1}{jk^j} = 
\frac{n}{k} + \frac{\log\left(\frac{k}{k-1}\right)}{k}$$
where we have used one of the series for $\log$ given here. The series is valid since $k \geq 2$.
Thus
$$\frac{n}{k} < \sum_{s=1}^{n}\operatorname{Li}_{s}\left(\frac{1}{k}\right) < \frac{n}{k} + \frac{\log(k) - \log(k-1)}{k}$$
Now for $n$ and $k$ natural numbers, the smallest possible difference between
$\left \lfloor \frac{n}{k}\right \rfloor + 1$ and $\frac{n}{k}$ is $\frac{1}{k}$.
Thus the result will hold if
$$\frac{\log(k) - \log(k-1)}{k} < \frac{1}{k}$$
or in other words if
$$\log(k) - \log(k-1) < 1$$
It's easy to check that $\log(x) - \log(x-1)$ is monotonically decreasing for $x > 1$. Also $\log(2) - \log(1) = \log(2) < 1$. It follows that the result holds for all natural numbers $k \geq 2$.
A: Some observaions:
For the case $k=1$ we have $$\sum_{s=1}^n\operatorname{Li}_s(1)=\sum_{s=1}^n\zeta(s)$$
Which is undefined unless you take $\zeta(1)=-1/12$. 
However through checking through Mathematica it is apparent that ignoring or extending $\zeta(1)$ gives the same answer which happens to be
$$\left\lfloor\sum_{s=1}^n\operatorname{Li}_s(1)\right\rfloor\stackrel{?}{=}n-1$$
A: For this proof to work, we have to add the restriction that $k \geq 2$, since $\operatorname{Li}_1\left( \frac{1}{1} \right)$ doesn't even converge anyway.

Now, by the definition of the polylogarithm, we have: 
\begin{align}
\sum_{s=1}^n \operatorname{Li}_s\left( \frac{1}{k} \right)
&=\sum_{s=1}^n \sum_{j=1}^\infty {1 \over j^s k^j}
\\&=\sum_{j=1}^\infty \frac{1}{k^j} \sum_{s=1}^n \frac{1}{j^s}
\\&=\frac{n}{k}+\sum_{j=2}^\infty \frac{1}{k^j} \left( {1-j^{-n} \over j-1} \right)
\end{align}

Now, let $R=\sum_{j=2}^\infty \frac{1}{k^j} \left( {1-j^{-n} \over j-1} \right)$, then obviously $R>0$. But also:
\begin{align}
R
&=\sum_{j=2}^\infty \frac{1}{k^j} \left( {1-j^{-n} \over j-1} \right)
\\&\leq \sum_{j=2}^\infty \frac{1}{k^j} \left( {1 \over j-1} \right)
\\&= \frac{1}{k} \sum_{j=1}^\infty {1 \over j k^j}
= {-\log \left(1-\frac{1}{k} \right) \over k}
\\&\leq {-\log \left(1-\frac{1}{2} \right) \over k}
={\log 2 \over k}
\\&< \frac{1}{k}
\end{align}
So that $0<R<\frac{1}{k}$.

Write $n=ak+b$, where $a,b \in \Bbb{Z}$, $a \geq 0$ and $0 \leq b \leq k-1$. Then we have:
$\displaystyle \quad \; {n \over k} = a+{b \over k} \\ \Rightarrow a \leq {n \over k} \leq a+{k-1 \over k} \\ \Rightarrow \left\lfloor {n \over k} \right\rfloor = a$. 
And we also have:
$\displaystyle \quad \; a+R \leq {n \over k}+R \leq a+{k-1 \over k}+R \\ \Rightarrow a<a+R \leq {n \over k}+R \leq a+{k-1 \over k}+R<a+{k-1 \over k}+{1 \over k} \\ \Rightarrow a< {n \over k}+R<a+1 \\ \Rightarrow \left \lfloor {n \over k}+R \right \rfloor = a$

Finally, since $\sum_{s=1}^n \operatorname{Li}_s\left( \frac{1}{k} \right)={n \over k}+R$, we have $$\left \lfloor \sum_{s=1}^n \operatorname{Li}_s\left( \frac{1}{k} \right) \right \rfloor = a = \left \lfloor {n \over k} \right \rfloor$$
