Evaluating a power series 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.
 A: 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 $f(1),\dots,f(a)$.
A: 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. 
A: 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 $(2.5^{-1}+1.5^{-2})/((1-5^{-2})3) = (2.5+1.1)/(24.3) = 11/72$.
