I'm currently interested in sums involving digit-functions. Especially, I'd like to calculate the following sum: $$ s=\sum_{k=1}^{\infty}{\frac{a_0(k)}{k\left(k+1\right)}} $$ Where $a_0(k)$ is the number of zeros in the base $10$ representation of $k$.
Normally, such sums can be tackled by considering products of the form $P(x,q)=\prod_{k=0}^{\infty}{\left(1+f_1(x)q^{10^k}+…+f_9(x)q^{9\cdot 10^k}\right)}$, but this approach seems not to work here, because the zero is a bit problematic.
Numerically, it seems that $s=\frac{10+\gamma+\psi{(0.1)}}{9}$ where $\psi$ is the digamma function.
Let $e(n)$ be the number of non-zero digits of $n$ then: $a_0(k)=\lfloor \log_{10}(k)\rfloor - e(k) $ so the idea is to compute: $$s_1=\sum_{k=1}^{\infty}{\frac{\lfloor \log_{10}(k)\rfloor}{k\left(k+1\right)}}\quad \quad s_2=\sum_{ k=1}^{\infty}{\frac{e(k)}{k\left(k+1\right)}}$$
The second sum is comptable by considering: $$P(x,q)=\prod_{k=0}^{\infty}{\left(1+x\left(q^{10^k}+…+q^{9\cdot 10^k}\right)\right)}$$ For the first a shifting using (in order tor replace the $\log_{10}(k)$ by a variable $n$) would produce :
$$s_2=\sum_{k=1}^{\infty}{\frac{\lfloor \log_{10}(k)\rfloor}{k\left(k+1\right)}}=\sum_{n=0}^{\infty}n\sum_{k=10^n}^{10^{n+1}-1}{\frac{1}{k\left(k+1\right)}}=\sum_{n=0}^{\infty} \left(\frac{n}{10^n}-\frac{n}{10^{n+1}}\right)$$
I hope that this is helpful.