Infinite Sum with Base Representations

Let $p = 2017$ . We will use the following lemmas:

Lemma 1 $v_p(n!) = \frac{n - s(n)}{p-1}$ Lemma 2 For $s > 1$ , $\frac{s}{s-1} > \zeta(s) > \frac{1}{s-1}$ The lemma follows from comparing $\zeta(s)$ to the area under the curve $f(t) = t^{-s}$ , giving us $1 + \int_1^\infty t^{-s} \, dt > \zeta(s) > \int_1^\infty t^{-s}.$ Corollary $\lim_{s \to 1^+} (s-1)\zeta(s) = 1$ .

We come back to the main problem. Observe that $\begin{aligned} \sum_{n \geq 1} \frac{s(n)}{n(n+1)} & = \sum_{n \geq 1} \frac{s(n)}{n} - \frac{s(n)}{n+1}\\ & = \sum_{n = 1}^\infty \frac{s(n) - s(n-1)}{n} \\ & = \sum_{n = 1}^\infty \frac{1 - (p-1)v_p(n)}{n}. \end{aligned}$

We define the function $F(s) = \sum_{n = 1}^\infty \frac{1 - (p-1)v_p(n)}{n^s}$ for $s > 1$ . This simplifies to $\begin{aligned} F(s) & = \zeta(s) - (p-1)\sum_{n = 1}^\infty \frac{v_p(n)}{n^s} \\ & = \zeta(s) - (p-1) \sum_{k = 1}^\infty \sum_{m = 1}^\infty \frac{1}{(p^km)^s} \\ & = \zeta(s) - (p-1) \zeta(s) \left ( \frac{p^{-s}}{1 - p^{-s}} \right ) \\ & = \zeta(s) \left ( \frac{p^s - p}{p^s - 1} \right ). \end{aligned}$ Now, we take the limit as $s \to 1^+$ , $\lim_{s \to 1^+} F(s) = \frac{1}{p-1}\lim_{s \to 1^+} \zeta(s)(s-1) \frac{p^s - p}{s-1} = \frac{p}{p-1} \lim_{s \to 1^+} \frac{p^{s-1} - 1}{s-1} = \frac{p}{p-1} \log p$ where the last equality follows from L'Hospital's Rule.

Let $s_p(n)$ be the sum of the digits of $n$ in the base $p$ representation of $n$ . Note that $s_p(n) = O(\ln n)$ as $m \to \infty$ , and hence it is clear that the infinite serie $S_p \; = \; \sum_{n \ge 1}\frac{s_p(n)}{n(n+1)}$ converges.

Let $j_p(n)$ be the index of $p$ in $n$ . It is easy to see that $s_p(n) - s_p(n-1) \; = \; 1 - (p-1)j_p(n) \hspace{2cm} n \ge 1$ and so, since $\sum_{n=1}^\infty \frac{s_p(n)}{n(n+1)} \; = \; \sum_{n \ge 1} \frac{s(n) - s(n-1)}{n} \; = \; \sum_{n \ge 1} \frac{1 - (p-1)j_p(n)}{n}$ we see that $n^{-1}$ gets added to the sum once for each $n$ , but $(p-1)n^{-1}$ is subtracted from the sum once each time $p$ divides $n$ . Thus $\sum_{n=1}^{p^K} \frac{s_p(n)}{n(n+1)} \; = \; \sum_{n=1}^{p^K} \frac{1}{n} - (p-1)\sum_{n=1}^{p^{K-1}}\frac{1}{pn} - (p-1)\sum_{n=1}^{p^{K-2}}\frac{1}{p^2n} - \cdots - (p-1)\sum_{n=1}^1 \frac{1}{p^Kn} \; = \; H_{p^K} - (p-1)\sum_{j=1}^K\frac{H_{p^{K-j}}}{p^j}$ Now we use the fact that $X_n \; = \; H_n - \ln n - \gamma \; = \; O\big(n^{-1}\big) \hspace{2cm} n \to \infty$ and calculate $\begin{aligned} \ln(p^K) - (p-1)\sum_{j=1}^K \frac{\ln(p^{K-j})}{p^j} & = \; \tfrac{p}{p-1}(1 - p^{-K})\ln p \\ 1 - (p-1)\sum_{j=1}^Kp^{-j} & = \; p^{-K} \end{aligned}$ so that $\sum_{n=1}^{p^K} \frac{s_p(n)}{n(n+1)} \; = \; \tfrac{p}{p-1}(1 - p^{-K})\ln p + \gamma p^{-K} + \left[ X_{p^K} - (p-1)\sum_{j=1}^K\frac{X_{p^{K-j}}}{p^j}\right]$ Since $|X_n| \le An^{-1}$ for some constant $A$ , we deduce that $\begin{aligned} \left| \sum_{n=1}^{p^K} \frac{s_p(n)}{n(n+1)} - \tfrac{p}{p-1}\ln p\right| & \le \; \left[\tfrac{p}{p-1}\ln p + \gamma + A\right]p^{-K} + (p-1)\sum_{j=1}^K\frac{A}{p^{K-j}\times p^j} \\ & \le \; \left[\tfrac{p}{p-1}\ln p + \gamma + A + AK(p-1)\right]p^{-K} \end{aligned}$ for all integers $K$ . Thus we deduce that $S_p \; = \; \tfrac{p}{p-1}\ln p$ . With $p=2017$ , we obtain the answer $\boxed{7.613141025}$ .