<s>[INST] 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.
 [/INST] 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 $$\left \lfloor \sum_{s=1}^n \operatorname{Li}_s\left( \frac{1}{k} \right) \right \rfloor = a = \left \lfloor {n \over k} \right \rfloor$$
</s>