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In the first chapter of Gouvea's intro to $p$-adics, there's a heuristic argument that

$$ \frac{2}{1}+\frac{2^2}{2}+\frac{2^3}{3}+\frac{2^4}{4}+\cdots=0 \tag{$\ast$}$$

as $2$-adic numbers, since it's the Mercator series for $\ln(-1)$ and $2\ln(-1)=\ln(-1)^2=\ln1=0$.

(Like I said, heuristic.)

I assume $(\ast)$ can be proven by analyzing $\ln(1+x)$ as a function of $p$-adic numbers, but there's an exercise that says we can show $(\ast)$ by elementary means. But how? I feel like I've considered this question in the past before, but don't remember if I ever solved it.

We should be able to prove it's congruent to $0$ mod $2^N$ for any $N$. This automatically truncates the series to a finite sum, and all of the denominators divisible by $2$ are underneath numerators even more divisible by $2$, so it's well-defined. Perhaps we can split the sum into subsums of even and odd indices and establish a recursion?

anon
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  • I don't understand what is left to prove after you have proved "it" (the limit, I assume?) is congruent to $0$ mod $2^N$ for every $N$. – Torsten Schoeneberg Jun 11 '18 at 20:03
  • Or are you asking: "What is an elementary proof for the statement For every $N \in \Bbb{N}$, there is $n \in \Bbb{N}$ such that $\sum_{1}^{m} \frac{2^i}{i} \equiv 0$ mod $2^N$ for all $m \ge n$"? – Torsten Schoeneberg Jun 11 '18 at 20:10
  • See OEIS sequence A087910 "Exponent of the greatest power of 2 dividing the numerator of 2^1/1 + 2^2/2 + 2^3/3 + ... + 2^n/n" which contains comment "asked for a proof that a(n) tends to infinity with n. While this is immediate from the theory of the 2-adic logarithm, elementary proofs are available". The link to Problems Competition leads to that proof. – Somos Jun 11 '18 at 20:11
  • One proves the series (the limit of the partial sums in the $2$-adic numbers) is $0$ by proving the statement in your second comment @Torsten. – anon Jun 11 '18 at 20:12

2 Answers2

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Since $\lim_{n\to\infty}\nu_2\left(\frac{2^n}n\right)=\infty$, it suffices to prove that there are partial sums with arbitrarily high $\nu_2$. We adapt the first solution from here (pp. 9–10).

For $n$ even, we know from the Binomial Theorem that $$1=(-1)^n=(1-2)^n=\sum_{k=0}^n\binom nk(-2)^k.$$ This implies \begin{equation}\sum_{k=1}^n\frac 1n\binom nk(-2)^k=0\label{1}\tag{1}.\end{equation} Now, we can write $$\frac1n\binom nk=\frac{(n-1)\cdots(n-(k-1))}{k!}=\frac{(-1)^{k-1}(k-1)!+nm_{n,k}}{k!}=\frac{(-1)^{k-1}}k+n\frac{m_{n,k}}{k!},$$ for some integer $m_{n,k}$. We can therefore rewrite $(\ref{1})$ as \begin{equation}\sum_{k=1}^n\frac{2^k}k=n\sum_{k=1}^n\frac{m_{n,k}(-1)^k2^k}{k!}.\label{2}\tag{2}\end{equation} It is well-known that $\nu_2(k!)<k$: this is easy to prove using Legendre's Formula. Therefore, the $\nu_2$ of the RHS of $(\ref{2})$ is at least the $\nu_2$ of $n$. In particular, $$\nu_2\left(\sum_{k=1}^{2^n}\frac{2^k}k\right)\geq n,$$ and we're done. $\blacksquare$

ViHdzP
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Like you said $$-\log(1-x)=\sum_{n\ge 1} \frac{x^n}{n}\in \Bbb{Q}[[x]]$$ converges in $\Bbb{Z}_2$ for $x\in 2\Bbb{Z}_2$ and the formal series in $\Bbb{Q}[[x]]$ satisfies $$(-2\log(1-x))'= \frac{2}{1-x}=\frac{(1-(1-x)^2)'}{1-(1-(1-x)^2)}=(-\log(1-x)^2)'$$ $$\implies -2\log(1-x)=-\log(1-x)^2$$ This relation of formal series stays true as a function on $2\Bbb{Z}_2$ so that $$-\log(1-2)=-\frac12 \log (1-2)^2=-\frac12 \log 1= 0$$

reuns
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  • I’m really not sure if this counts as “elementary”. OP words the question as if they wanted to avoid using $p$-adic number theory in the first place. – ViHdzP Jan 06 '20 at 04:55
  • It is elementary. If you want to avoid formal series then reduce $-\log(1-2x)$ modulo $\bmod (2^N,(2x)^N)$ it doesn't change the argument, you'll obtain that $\log (-1)$ is $0 \bmod 2^{N-\log N}$. – reuns Jan 06 '20 at 04:57