Does $\sum\limits_{n=1}^\infty\frac{1}{\text{Sum of permutations of digits of }n}$ converge?

Question

Hopefully the following chart explains some things: $$\begin{array}{|l|l|} \hline n & \frac{1}{\text{Sum of permutations of digits of }n} \\ \hline 1 & \frac{1}{1} \\ \hline 2 & \frac{1}{2} \\ \hline 3 & \frac{1}{3} \\ \hline 10 & \frac{1}{10+01} \\ \hline 11 & \frac{1}{11+11} \\ \hline 172 & \frac{1}{172+127+271+217+712+721} \\ \hline \end{array}$$

I know the harmonic series diverges but I can't compare it to that because it is bigger after $n=9$. How can I test whether or not it converges?

This problem is in base $10$ but it would be interesting to see it generalized to all bases.

Answer 1

Let's discuss the sum of all digit permutations of the numbers that have $n$ digits. There are $n!$ premutations, meaning that the denominator is AT LEAST $n!$. If we write the sum this way: \begin{equation} \sum^{\infty}_{n=1,n-number\ of\ digits}\frac{1}{sum\ of\ permutations}\leq \sum^{\infty}_{n=1,n-number\ of\ digits}\frac{1}{n!} \end{equation} There are at most $10^n$ (not precisely, but does not affect convergence) numbers that have $n$ digits, meaning: \begin{equation} \sum^{\infty}_{n=1,n-number\ of\ digits}\frac{1}{n!}\leq \sum^{\infty}_{n=1}\frac{10^n}{n!} \end{equation} The last sum converges (easily provable by some of the criteria), meaning that your sum converges too. The key was to think in terms of number of digits rather than the numbers themselves.

Answer 2

Let $\mathcal{W}_d = \{1, \ldots, 9\} \times \{0, \ldots, 9\}^{d-1}$ denote the set of all strings for the $d$-digit decimal expressions. If we denote OP's sum by $S$, then

\begin{align*} S &= \sum_{d=1}^{\infty} \sum_{(a_1,\ldots,a_d) \in \mathcal{W}_d} \frac{1}{(d-1)!(\underbrace{11\ldots11}_{d})(a_1 + \cdots + a_d)} \\ &= \sum_{d=1}^{\infty} \frac{9}{(d-1)!(10^d - 1)} \sum_{(a_1,\ldots,a_d) \in \mathcal{W}_d} \frac{1}{a_1 + \cdots + a_d}. \end{align*}

Clearly $|\mathcal{W}_d| = 9 \cdot 10^{d-1}$, and so, $S$ is bounded by

$$ S \leq \sum_{d=1}^{\infty} \frac{9}{(d-1)!(10^d - 1)} |\mathcal{W}_d| < \infty. $$