Sum of the series $\sum_{n=0}^{\infty} \lfloor nr \rfloor x^n$ where $r$ is rational?

Can we find the exact sum of the series $$\sum_{n=0}^{\infty} \lfloor nr \rfloor x^n$$ where $$r$$ is rational?

There is a special case given here but I don't know how to prove it and can we get the sum for the general case?

• take a look here – Masacroso Nov 23 '19 at 17:59
• @Masacroso Your link deals with finite sum. I found this though containg infinite sum. – mathisgood Nov 23 '19 at 18:14
• If $r = a/b$ in lowest form, then for any positive integer $k$, $\lfloor (n + kb)r\rfloor = \lfloor nr \rfloor +ka$. This can be used to reduce to finding the sum of the first $b$ terms of the sum. I'm not sure if there's a nice formula for that in general, however. – Zarrax Nov 23 '19 at 19:05

First, this is not a full answer but enough significant progress that I felt it was worth typing up.

Note that the sum only converges if $$|x|<1$$. As such, for the remainder of this post we shall assume $$-1.

Let $$r=a/b$$ be rational. If $$r=a/b$$ is an integer, then the sum is easily computed to be

$$\sum_{n=0}^\infty\left\lfloor n \frac{a}{b}\right\rfloor x^n=\sum_{n=0}^\infty n rx^n=\frac{r x}{(x-1)^2}$$

Thus, we may as well assume $$r=a/b$$ is not an integer from here on out. Now, we can write $$n$$ as $$n=bq+s$$ where $$q\in\{0,1,2,3,...\}$$ and $$0\leq s . Then

$$\left\lfloor n \frac{a}{b}\right\rfloor=\left\lfloor (bq+s) \frac{a}{b}\right\rfloor=qa+\left\lfloor s \frac{a}{b}\right\rfloor$$

Then the sum can be rewritten as

$$\sum_{n=0}^\infty \left\lfloor n \frac{a}{b}\right\rfloor x^n=\sum_{q=0}^\infty\sum_{s=0}^{b-1}\left\lfloor (bq+s) \frac{a}{b}\right\rfloor x^{bq+s}=\sum_{q=0}^\infty\sum_{s=0}^{b-1}\left(qa+\left\lfloor s \frac{a}{b}\right\rfloor\right) x^{bq+s}$$

We can split this sum into two parts:

$$\sum_{q=0}^\infty\sum_{s=0}^{b-1}\left(qa+\left\lfloor s \frac{a}{b}\right\rfloor\right) x^{bq+s}=\sum_{q=0}^\infty\sum_{s=0}^{b-1}qa x^{bq+s}+\sum_{q=0}^\infty\sum_{s=0}^{b-1}\left\lfloor s \frac{a}{b}\right\rfloor x^{bq+s}$$

The first infinite sum is easily found to be

$$\sum_{q=0}^\infty\sum_{s=0}^{b-1}qa x^{bq+s}=\frac{a x^b}{(x-1) \left(x^b-1\right)}$$

Since the second infinite sum converges absolutely, we may as well sum the $$q$$ sum before the $$s$$ sum. That is

$$\sum_{q=0}^\infty\sum_{s=0}^{b-1}\left\lfloor s \frac{a}{b}\right\rfloor x^{bq+s}=\sum_{s=0}^{b-1}\sum_{q=0}^\infty\left\lfloor s \frac{a}{b}\right\rfloor x^{bq+s}$$

$$=\sum_{s=0}^{b-1}\left(\left\lfloor s \frac{a}{b}\right\rfloor x^s\sum_{q=0}^\infty x^{bq}\right)=\sum_{s=0}^{b-1}\left(\left\lfloor s \frac{a}{b}\right\rfloor x^s\frac{1}{1-x^b}\right)$$

$$=\frac{1}{1-x^b}\sum_{s=0}^{b-1}\left\lfloor s \frac{a}{b}\right\rfloor x^s$$

The question then becomes: what is the sum

$$\sum_{s=0}^{b-1}\left\lfloor s \frac{a}{b}\right\rfloor x^s?$$

Now, we can slightly simplify this if we assume $$a/b$$ is in its most reduced form. That is, $$\gcd(a,b)=1$$ (if $$a=0$$, then the original sum is clearly $$0$$ so we may ignore this case). We may then write $$a=mb+t$$ for $$m\in\mathbb{Z}$$, $$\gcd(t,b)=1$$, and $$0< t. Note that if $$t=0$$, then $$a/b$$ is an integer. Otherwise, the sum is

$$\sum_{s=0}^{b-1}\left\lfloor s \frac{a}{b}\right\rfloor x^s=\sum_{s=0}^{b-1}\left\lfloor s \frac{mb+t}{b}\right\rfloor x^s=\sum_{s=0}^{b-1}\left\lfloor s \frac{mb+t}{b}\right\rfloor x^s$$

$$=\sum_{s=0}^{b-1}smx^s+\sum_{s=0}^{b-1}\left\lfloor s \frac{t}{b}\right\rfloor x^s=\frac{m \left(b x^{b+1}-x^{b+1}-b x^b+x\right)}{(x-1)^2}+\sum_{s=0}^{b-1}\left\lfloor s \frac{t}{b}\right\rfloor x^s$$

We have now further simplified the question to: what is the sum

$$\sum_{s=0}^{b-1}\left\lfloor s \frac{t}{b}\right\rfloor x^s$$

where $$\gcd(t,b)=1$$ and $$0? Unfortunately, this seems like a difficult problem. Here are the first few such sums, for $$b=1,2,\cdots ,6$$:

$$\left( \begin{array}{c} \{0\} \\ \{0\} \\ \left\{0,x^2\right\} \\ \left\{0,2 x^3+x^2\right\} \\ \left\{0,x^4+x^3,2 x^4+x^3+x^2,3 x^4+2 x^3+x^2\right\} \\ \left\{0,4 x^5+3 x^4+2 x^3+x^2\right\} \\ \end{array} \right)$$

Overall, the sum is

$$\frac{a x^b}{(x-1) \left(x^b-1\right)}+\frac{1}{1-x^b}\left( \frac{m \left(b x^{b+1}-x^{b+1}-b x^b+x\right)}{(x-1)^2}+\sum_{s=0}^{b-1}\left\lfloor s \frac{t}{b}\right\rfloor x^s \right)$$

where $$m=\lfloor a/b\rfloor$$, $$\gcd(t,b)=1$$, and $$0.

Just to show that this answer is correct, we can rederive the special case given in the link above. In this case, $$a=1$$, implying the second term is $$0$$ as

$$m=\lfloor a/b\rfloor=\lfloor 1/b\rfloor=0$$

and

$$\sum_{s=0}^{b-1}\left\lfloor s \frac{t}{b}\right\rfloor x^s =\sum_{s=0}^{b-1}\left\lfloor s \frac{1}{b}\right\rfloor x^s =\sum_{s=0}^{b-1}\left\lfloor \frac{s}{b}\right\rfloor x^s =0$$

as $$0\leq s. Thus, the sum is simply

$$\frac{x^b}{(x-1) \left(x^b-1\right)}.$$

In the example, they use $$k=1/x$$, which gives us

$$\frac{ (1/k)^b}{((1/k)-1) \left((1/k)^b-1\right)}=\frac{ (1/k)^bk^{b+1}}{((1/k)-1) \left((1/k)^b-1\right)k^{b+1}}=\frac{k}{(k-1)(k^b-1)}$$

which is what they got.