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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.

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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 $k \geq 2$.

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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 $$\left\lfloor\sum_{s=1}^n\operatorname{Li}_s(1)\right\rfloor\stackrel{?}{=}n-1$$

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  • $\begingroup$ @user153012 As I said it doesn't matter. $\endgroup$ – Ali Caglayan Sep 29 '14 at 16:44
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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 $$\left \lfloor \sum_{s=1}^n \operatorname{Li}_s\left( \frac{1}{k} \right) \right \rfloor = a = \left \lfloor {n \over k} \right \rfloor$$

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  • $\begingroup$ Oh hell. I wrote this answer thinking no body responded because I left the page unrefreshed from yesterday in my browser. If this is the same as some other proof that was posted, I apologize. $\endgroup$ – Fujoyaki Sep 29 '14 at 22:46
  • $\begingroup$ No problem, thank you for your proof. $\endgroup$ – user153012 Sep 29 '14 at 23:19

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