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Can somone help me find a closed form expression for this sum given any rational value of x, and any integer p, where {x} denotates the fractional part of x.$$\sum_ {k=1}^{\infty}\frac{ \left\{p^kx \right\} }{p^k} $$ It also seems to converge to rational values.

For example if i let $p=5,x=\frac13$, the series converges to $\frac{11}{72}$

I don't think it should be very hard, but im not sure, I would appreciate any help.

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I think this shouldn't be possible for any $a,b \in \mathbb{Z}$, for example, if $b = 0$ and $a = -1$, we're trying to find $\sum_k f(-k)$, about which we have no information. –  uncookedfalcon Nov 24 '12 at 8:22
    
I changed the question slightly, your comment doesn't really apply in this case. –  Ethan Nov 30 '12 at 9:12
    
I think I can do it for rational x –  Amr Nov 30 '12 at 9:20
    
There is no need to put a notification in the title. Titles are for titles, not for notifications. –  Asaf Karagila Nov 30 '12 at 9:35
    
please do so, thanks alot –  Ethan Nov 30 '12 at 9:42
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4 Answers 4

up vote 4 down vote accepted
+100

Suppose $x\in\mathbb{Q}$, $0\lt x\lt1$, and $x$ has the base-$p$ expansion $$ x=\sum_{k=1}^\infty\frac{d_k}{p^k}\tag{1} $$ Then $$ \frac{\{p^nx\}}{p^n}=\sum_{k=n+1}^\infty\frac{d_k}{p^k}\tag{2} $$ So that $$ \begin{align} f_p(x) &=\sum_{n=0}^\infty\frac{\{p^nx\}}{p^n}\\ &=\sum_{n=0}^\infty\sum_{k=n+1}^\infty\frac{d_k}{p^k}\\ &=\sum_{k=1}^\infty\sum_{n=0}^{k-1}\frac{d_k}{p^k}\\ &=\sum_{k=1}^\infty\frac{k\,d_k}{p^k}\tag{3} \end{align} $$ Since the sum in $(3)$ starts at $k=0$, $f_p(x)-x$ is the function in the question. However, if $f_p(x):\mathbb{Q}\mapsto\mathbb{Q}$, then $f_p(x)-x:\mathbb{Q}\mapsto\mathbb{Q}$.

Finite base-$p$ expansion

Obviously, if the base-$p$ expansion of $x$ is finite, then the sum in $(3)$ is finite $$ f_p(x)=\sum_{k=1}^m\frac{k\,d_k}{p^k}\tag{4} $$ which is a finite sum of rational numbers, hence $f_p(x)\in\mathbb{Q}$.

Repeating base-$p$ expansion

If the base-$p$ expansion of $x$ repeats with period $m$, then $$ \begin{align} f_p(x) &=\sum_{k=1}^m\sum_{n=0}^\infty\frac{(k+nm)d_k}{p^{k+nm}}\\[6pt] &=\sum_{k=1}^m\frac{d_k}{p^k}\sum_{n=0}^\infty\frac{k+nm}{p^{nm}}\\[6pt] &=\sum_{k=1}^m\frac{d_k}{p^k}\left(\frac{kp^m}{p^m-1}+\frac{mp^m}{(p^m-1)^2}\right)\\[6pt] &=\frac1{p^m-1}\left(mx+p^m\sum_{k=1}^m\frac{k\,d_k}{p^k}\right)\tag{5} \end{align} $$ which is again a finite sum of rational numbers, hence $f_p(x)\in\mathbb{Q}$.

Mixed base-$p$ expansions

Note that if there are no base-$p$ carries when adding $x$ and $y$, then each digit of the sum is the sum of the digits, and therefore, by $(1)$ and $(3)$, $$ f_p(x+y)=f_p(x)+f_p(y)\tag{6} $$ Furthermore, $$ \begin{align} f_p\left(\frac{x}{p^n}\right) &=\sum_{k=1}^\infty\frac{(k+n)d_k}{p^{k+n}}\\ &=\frac1{p^n}\left(nx+f_p(x)\right)\tag{7} \end{align} $$ Combining $(4)$, $(5)$, $(6)$, and $(7)$, we get

Conclusion

If $x\in\mathbb{Q}$, then $$ \sum_{k=1}^\infty\frac{\{p^kx\}}{p^k}=f_p(x)-x\in\mathbb{Q} $$


Example 1

In base $5$, $\frac{14}{25}=.\color{#C00000}{24}$. By $(4)$ $$ \begin{align} f_5\left(\frac{14}{25}\right) &=\frac{\color{#00A000}{1}\cdot\color{#C00000}{2}}{5^{\color{#00A000}{1}}}+\frac{\color{#00A000}{2}\cdot\color{#C00000}{4}}{5^{\color{#00A000}{2}}}\\[6pt] &=\frac{18}{25} \end{align} $$

Example 2

In base $5$, $\color{#0000FF}{\frac13}=.\overline{\color{#C00000}{13}}$, therefore, $p=5,m=2,d_1=1,d_2=3$. By $(5)$ $$ \begin{align} f_5\left(\color{#0000FF}{\frac13}\right) &=\frac1{5^2-1}\left(2\cdot\color{#0000FF}{\frac13}+5^2\left(\frac{\color{#00A000}{1}\cdot\color{#C00000}{1}}{5^\color{#00A000}{1}}+\frac{\color{#00A000}{2}\cdot\color{#C00000}{3}}{5^\color{#00A000}{2}}\right)\right)\\[6pt] &=\frac{35}{72} \end{align} $$

Example 3

In base $5$, $\frac{67}{75}=.24\overline{13}$. Using $(6)$ and $(7)$ and the previous examples, we get $$ \begin{align} f_5\left(\frac{67}{75}\right) &=f_5\left(\frac{14}{25}\right)+f_5\left(\frac13\cdot\frac1{25}\right)\\[6pt] &=\frac{18}{25}+\frac1{5^2}\left(2\cdot\frac13+\frac{35}{72}\right)\\[6pt] &=\frac{1379}{1800} \end{align} $$

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I found the way you related this expression to the base p expansion of x, very clever. And I appreciate all the work you spent, but is there a way I can give the sum involving the weighted terms in base p of x, in a closed form expression, with out having to re-write x in base p? –  Ethan Nov 30 '12 at 19:03
    
If you are asking if there is a way of writing $$ \sum_{k=1}^\infty\frac{k\,d_k}{p^k} $$ given only $$ x=\sum_{k=1}^\infty\frac{d_k}{p^k} $$ I don't think so, other than $$ \sum_{k=1}^\infty\frac{\{p^kx\}}{p^k} $$ –  robjohn Nov 30 '12 at 19:44
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For many pairs $(a,b)$, you can take linear combinations of $g(1/a),\dots,g(a/a)$ to obtain the sum in question. For example, with $a=6$, \begin{align*} -g(\tfrac x6) + g(\tfrac{3x}6) + g(\tfrac{4x}6) &= 6 \sum_{k=0}^\infty f(6k+1) \\ g(\tfrac{2x}6) - g(\tfrac{3x}6) - g(\tfrac{4x}6) + g(\tfrac{5x}6) &= 6 \sum_{k=0}^\infty f(6k+2) \\ g(\tfrac{x}6) - g(\tfrac{2x}6) - g(\tfrac{4x}6) + g(\tfrac{5x}6) &= 6 \sum_{k=0}^\infty f(6k+3) \\ g(\tfrac{x}6) - g(\tfrac{2x}6) - g(\tfrac{3x}6) + g(\tfrac{4x}6) &= 6 \sum_{k=0}^\infty f(6k+4) \\ g(\tfrac{2x}6) + g(\tfrac{3x}6) - g(\tfrac{5x}6) &= 6 \sum_{k=0}^\infty f(6k+5). \end{align*} Finding the exact linear combinations is a simple linear algebra problem for any $a$: since everything in sight is period with period $a$, you only need to get the right coefficients of $f(1),\dots,f(a)$.

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I altered the question slightly before I made it into a bounty, your work is similar to the original question, but in this exact sense no longer applies. –  Ethan Nov 26 '12 at 6:56
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Let $x=r/s$, then $p^k$ is eventually periodic modulo $s$, so $\{{p^kx\}}$ is eventually periodic, so the series is eventually geometric with common ratio some inverse power of $p$, so the sum is rational.

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let $x = a/b$ with $b>0$ and $a$ coprime with $b$.

The fractional part of $p^ka/b$ only depends on the value of $p^ka \pmod b$, and the sequence $(p^ka \pmod b)_{k \ge 0}$ is eventually periodic : Since there are $b$ classes modulo $b$, when you look at $p^1a, p^2a, \ldots p^{b+1}a \pmod b$, you will find two exponents $k_0,k_1$ such that $p^{k_0}a = p^{k_1}a \pmod b$. And since the remainder of $px \pmod b$ only depends of $x \pmod b$, from that point on, $p^{k_1+k}a = p^{k_0+k}a \pmod b$, so the sequence is $T$-periodic with $T = k_1-k_0$.

Thus if you write $p^k a = a_k \pmod b$ where $0 \le a_k < b$, the sequence $(a_k)$ is eventually $T$-periodic, and $\{p^k x\} = a_k/b$, so $\sum_{k \ge 1} \{p^k x\}p^{-k} = \sum_{k \ge 0} a_k p^{-k} b^{-1}$.

Now, $( \sum_{k \ge 1} a_k p^{-k}b^{-1})(p^T - 1) = \sum_{k \ge 1} (a_{k-T} - a_k)p^{-k}b^{-1}$, where we put $a_k = 0$ for $k \le 0$ where necessary. Since $(a_k)$ is eventually $T$-periodic, $(a_{k-T} - a_k)$ only has finitely many nonzero terms (in fact it is zero at least when $k \ge k_1$), thus $( \sum_{k \ge 1} a_k p^{-k}b^{-1})(1 - p^{-T}) = \sum_{1 \le k < k_1} (a_k - a_{k-T})p^{-k}b^{-1}$

Putting everything back together, we get $\sum_{k \ge 1} \{p^k x\}p^{-k} = (\sum_{1 \le k < k_1} (a_k - a_{k-T})p^{-k})/((1 - p^{-T})b)$, which is rational.

For example with $p=5,a=1,b=3$ we have $p^1 a = p^3 a = 2 \pmod 3$ so we can pick $k_1 = 3$ and $T = 2$. We have $a_1 = 2, a_2 = 1$, so the formula gives $(2.5^{-1}+1.5^{-2})/((1-5^{-2})3) = (2.5+1.1)/(24.3) = 11/72$.

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