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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?

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  • $\begingroup$ take a look here $\endgroup$ – Masacroso Nov 23 '19 at 17:59
  • $\begingroup$ @Masacroso Your link deals with finite sum. I found this though containg infinite sum. $\endgroup$ – mathisgood Nov 23 '19 at 18:14
  • $\begingroup$ 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. $\endgroup$ – Zarrax Nov 23 '19 at 19:05
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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 $-1<x<1$.

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 $0\leq s <b$. 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 $0< t<b-1$. 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 $0<t<b$? 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 $0<t<b$.

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 $0\leq s<b$. 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.

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