Repetition in values of sequences of type $s_{n+1} = s_n + n^x / 10^n$

Question

Assume we have a sequence of type $a_n=\frac{n^x}{10^n}$.The corresponding series would be one with the formula $s_{n+1}=s_n+\frac{n^x}{10^n}$.

We can obviously see that when x = 1, this would result in $s_n$ approaching the value 0.123456... And indeed, when I checked using Python, I got the following output: 0.1234567901234567901234567901234567901234567901234567901234567901234567901234567901234567901234567901234567901234567901234567901234567901234567901234567901234567901234567901234567901234567901234567901234567901234567901234567901234567901234567901234567901234567901234567901234567901234567898

This matches the idea that there should be a repeating pattern. However, when instead using x = 2 or 3, I also get a repitition. These are my outputs: 0.1508916323731138545953360768175582990397805212620027434842249657064471879286694101508916323731138545953360768175582990397805212620027434842249657064471879286694101508916323731138545953360768175582990397805212620027434842249657064471879286694101508916323731138545953360768175582990397804271 (repitition always starts with 1508) and 0.2149062642889803383630544124371284865112025605852766346593507087334247828074988568815729309556470050297210791037951531778692272519433013260173754000914494741655235482395976223136716963877457704618198445358939186099679926840420667581161408321902149062642889803383630544124371284865111751489 (with the repitition always starting with 2149)

Where does this pattern stem from?

My code can be found here.

Answer 1

Consider the standard geometric series (defined over real numbers $y$ such that $|y|<1$): $$f(y)=\sum_{n=0}^{+ \infty} y^n= \frac{1}{1-y}$$ And compute the first derivative of $f$: $$f'(y)=\sum_{n=0}^{+ \infty} ny^{n-1} = \frac{1}{(1-y)^2} $$ so that $$yf'(y)=\sum_{n=0}^{+ \infty} ny^n = \frac{y}{(1-y)^2}$$ Now plug $y= 1/10$ to get $$\frac{1}{10} + \frac{2}{10^2} + \frac{3}{10^3} + \frac{4}{10^4} + \cdots = \sum_{n=0}^{+ \infty} n(1/10)^n = \frac{(1/10)}{(1-(1/10))^2}= \frac{10}{81}$$ whose decimal expansion is indeed $0.123456790123...$

To compute the values of the other series you need to involve derivatives of higher order, but in general you will get always rational functions. Plugging a rational number (i.e. $y=1/10$) will always give you a rational number (thus periodic decimal) as a result.

Answer 2

For $x\in \{0,1,2,...\}$, let$$ b_x = \sum_{n=1}^\infty \frac{n^x}{10^n} = \frac{1^x}{10^1} + \frac{2^x}{10^2} + \frac{3^x}{10^3} +\cdots $$ so $b_x$ is the limit of your sequence. Observe $b_0 = 0.1111...=\frac{1}{9}$. We will show that $b_x$ is always rational for non-negative integers $x$. Suppose $x\ge 1$: \begin{eqnarray} b_{x} &=& \frac{1^x}{10^1} + \frac{2^x}{10^2} + \frac{3^x}{10^3} +\cdots \\ &=& \frac{1}{10} + \frac{1}{10}\left(\frac{(1+1)^{x}}{10^1} + \frac{(1+2)^x}{10^2} + \frac{(1+3)^x}{10^3}+\cdots\right) \end{eqnarray} Expanding out each term using the binomial theorem, we obtain \begin{eqnarray} b_x &=& \frac{1}{10} + \frac{1}{10}\left(\frac{(1+1)^{x}}{10^1} + \frac{(1+2)^x}{10^2} + \frac{(1+3)^x}{10^3}+\cdots\right)\\ &=& \frac1{10} + \frac{1}{10}\left(\frac{\binom{x}{0}1^0 + \binom{x}{1}1^1 + \cdots + \binom{x}{x}1^x }{10^1} + \frac{\binom{x}{0}2^0 + \binom{x}{1}2^1 + \cdots + \binom{x}{x}2^x }{10^1} +\cdots\right) \end{eqnarray} grouping like terms together, we get \begin{eqnarray} b_x &=& \frac1{10} + \frac{1}{10}\left(\binom{x}{0}\left(\frac{1^0}{10^1} + \frac{2^0}{10^2} + \frac{3^0}{10^3} +\cdots\right) + \binom{x}{1}\left(\frac{1^1}{10^1} + \frac{2^1}{10^2} + \frac{3^1}{10^3} +\cdots\right) + \cdots\\ \cdots+ \binom{x}{x-1}\left(\frac{1^{x-1}}{10^1} + \frac{2^{x-1}}{10^2} + \frac{3^{x-1}}{10^3} +\cdots\right) + \binom{x}{x}\left(\frac{1^x}{10^1} + \frac{2^x}{10^2} + \frac{3^x}{10^3} +\cdots\right)\right) \\ &=&\frac1{10} + \frac{1}{10}\left(\binom{x}{0}b_0 + \binom{x}{1}b_1 + \cdots \binom{x}{x-1}b_{x-1} + \binom{x}{x}b_x\right) \end{eqnarray} Using the fact that $\binom{x}{x} = 1$, we can solve this equation for $b_x$ and obtain the recurrence relation $$ b_x = \frac19 + \frac19 \binom{x}{0}b_0 + \frac19\binom{x}{1}b_1 + \cdots + \frac19\binom{x}{x-1}b_{x-1} $$ Clearly, we will always get a rational number since each term is just a sum of rational numbers that we got previously in the sequence.

Computation of the first few terms:\begin{eqnarray} b_0 &=& \frac19\\ b_1 &=& \frac19 + \frac{1}9\binom{1}{0}b_0 = \frac{1}{9} + \frac{1}{9}\cdot\frac{1}{9} = \frac{10}{81} = 0.12345678901234...\\ b_2 &=& \frac19 + \frac19\binom{2}{0}b_0 + \frac19\binom{2}{1}b_1 = \frac19+\frac{1}{9}\cdot\frac19 + 2\cdot\frac19\cdot\frac{10}{81} = \frac{110}{729} \approx 0.15089...\\ b_3 &=& \frac19 + \frac{1}{9}\binom{3}{0}b_0 + \frac{1}{9}\binom{3}{1}b_1 + \frac{1}{9}\binom{3}{2}b_2 = \frac{470}{2187} \approx 0.21490...\\ b_4 &=&\frac19 + \frac{1}{9}\binom{4}{0}b_0 + \frac{1}{9}\binom{4}{1}b_1 + \frac{1}{9}\binom{4}{2}b_2 +\frac{1}{9}\binom{4}{3}b_3= \frac{7370}{19683} \approx 0.37443...\\ \vdots \end{eqnarray}

More generally, $f(x,y) = \sum_{n=1}^\infty \frac{n^x}{y^n}$ is always rational for nonnegative integers $x$ and rational numbers $y$. This is a special case of the polylogarithm with a negative index; $f(x,y) = \mathrm{Li}_{-x}(1/y)$. For $x\in\{0,1,2,...\}$, this is a rational function of $y$ (see Crostul's answer).