Combinatorial proof that 2-adic sum $\sum_{n>0} 2^n/n=0$

Question

My teacher gave me an idea that there should be an interesting combinatorial proof that the sums $$S_n := \sum_{i=1}^{n} \frac{2^i}{i}$$ are eventually divided by any power of $2$, that is $\lim_{n \to \infty} \operatorname{ord}_2 (S_n)=\infty$.

There is a short proof using 2-adic limit of $S_n$'s. The additional terms $2^i/i$ tend to zero 2-adically, so the 2-adic limit exists. But $$\sum_{i\geqslant 1} \frac{2^i}{i}=\log(1-2)=\log(-1)=\frac{\log((-1)^2)}{2}=0,$$ thus the sums tend to zero 2-adically.

Answer 1

$$S_n = \sum_{j=1}^{n}\frac{1}{j}\sum_{k=0}^{j}\binom{j}{k}=H_n+\sum_{1\leq k\leq j\leq n}\frac{1}{k}\binom{j-1}{k-1}=H_n+\sum_{k=1}^{n}\binom{n}{k}\frac{1}{k}$$ leads to: $$ S_n = \sum_{k=1}^{n}\binom{n}{k}\frac{1-(-1)^k}{k}=2\sum_{\substack{k\leq n \\ k\text{ odd}}}\binom{n}{k}\frac{1}{k} $$ that is pretty simple to study $2$-adically!
That is related with the nice identity $$ \sum_{k=0}^{n}\frac{(-1)^k}{2k+1}\binom{n}{k}=\int_{0}^{1}(1-x^2)^n\,dx = \frac{(2n)!!}{(2n+1)!!}=\frac{4^n n!^2}{(2n+1)!}$$ from which $$ \nu_2\left(\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}\binom{n}{k}\right)=2n+\sum_{h\geq 1}\left(2\left\lfloor\frac{n}{2^h}\right\rfloor-\left\lfloor\frac{2n+1}{2^h}\right\rfloor\right)\geq 2n-\log_2(n).$$

Answer 2

Indeed, more generally (and emphatically noncombinatorially), if $\zeta$ is a $p^n$-th root of unity, then the series for $\log(\zeta)$ is $p$-adically convergent, and must converge to zero. As, with $\log(1+z)=-\sum_{m\ge1}(-z)^m/m$, if you want the log of $i$, you substitute $i-1$ for $z$. Since $\text{ord}_2(i-1)=1/2>0 $, the series is convergent.

What makes this work is that the $p$-adic logarithmic series, defined for all $z$ with $\text{ord}_p(z)>0$, makes the logarithm a homomorphism, unlike the complex logarithm. That’s why the complex log of $i$ isn’t zero.