elementary proof of $\log(-1)=0$ in $\mathbb{Q}_2$

Question

I'm looking for an elementary proof of the fact that $$\sum_{n=1}^\infty\frac{2^n}{n}=0$$ in the field $\mathbb{Q}_2$ of $2$-adic numbers. I know that this follows from the functional equation of the $p$-adic logarithm $0=\log(1)=\log((-1)^2)=2\log(-1)$, but I like to see a more straight-forward argument.

Answer 1

Expanding @Merosity's idea, here is a solution: Let $N$ be a positive integer. Then

\begin{align*} 0 = \frac{1 - (1 - 2)^{2^N}}{2^N} &= \sum_{k=1}^{\infty} \frac{1}{2^N} \binom{2^N}{k} (-1)^{k-1} 2^k \\ &= \sum_{k=1}^{\infty} \left[ \prod_{l=1}^{k-1} \left( 1 - \frac{2^N}{l} \right) \right] \frac{2^k}{k}. \end{align*}

• In the second step, we utilized the binomial theorem.

• In the third step, we utilized the identity $\frac{1}{n}\binom{n}{k} = \frac{1}{k} \binom{n-1}{k-1} $

By the dominated convergence theorem, as $N \to \infty$ this converges to

$$ 0 = \sum_{k=1}^{\infty} \left[ \lim_{N\to\infty} \prod_{l=1}^{k-1} \left( 1 - \frac{2^N}{l} \right) \right] \frac{2^k}{k} = \sum_{k=1}^{\infty} \frac{2^k}{k} $$

and therefore the desired conclusion follows.