Resistance across opposite vertices of n-dimensional cube with each edge at one ohm resistance is $$R_n=\sum_{k=0}^{n-1}\frac1{(n-k){n\choose k}}=\frac1n\sum_{k=0}^{n-1}\frac1{{n-1\choose k}}.$$ The proof is simple: we can identify nodes with equal voltages, and there are ${n\choose k}$ nodes with a distance $k$ to a given point.
From another hand $$R_n=\frac1{2^n}\sum_{k=1}^{n}\frac{2^k}{k},$$ because both sums satisfy the same recurrence relation $$R_n=\frac1n+\frac12R_{n-1}.$$
We can consider the sum $\sum_{k=1}^{n}\frac{2^k}{k}$ as a partial sum of $p$-adic logarithm $$\log_p(1+x)=-\sum_{k=1}^{\infty}\frac{(-x)^k}{k}$$ at the point $x=2$ (it is well defined for $p=2$). Using two formulas for $R_n$ we can get a simple application. We can find the value of $2$-adic logarithm at the point $-2$: $$-\log_2(-2)=\lim_{n\to\infty}\sum_{k=1}^{n}\frac{2^k}{k}= \lim_{n\to\infty}\frac{2^n}n\sum_{k=0}^{n-1}\frac1{{n-1\choose k}}=0.$$ It is not a surprise (see § 4.4.11 from Cohen (2007), Number theory, Volume I: Tools and Diophantine equations).
First question: why does $2$-adic logarithm arise in combinatorial problem?
Second question: do you know any more connections between combinatorial and $p$-adic objects?
