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The estimate I wanna get is $$v_2(\sum_{i=1}^n\frac{2^i}{i})\geq\min_{t\geq n+1}\{t-v_2(t)\}\tag{*}$$ where $v_2$ is the highest power of 2 defined on $\mathbb{Q}$.

Set $$S_n:=\sum_{i=1}^n\frac{2^i}{i}.$$

One knows that $$\log(-1)=\log(1+(-2))\equiv -\sum_{i=1}^n\frac{2^i}{i}+\text{multiples of }2^{n+1}\equiv -\sum_{i=1}^n\frac{2^i}{i}\mod 2^{n+1}$$ so we have$$ 0=2\log(-1)\equiv-2S_n\mod 2^{n+1} $$ which means $$ v_2(2S_n)\geq n+1. $$

So we might get $v_2(S_n)\geq n$.

But I still get trouble showing the inequility (*).

Lucellia Kassel
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1 Answers1

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Remember the "ultrametric" inequality $v(a+b)\ge min(v(a),v(b))$ and, in case $v(a) \neq v(b)$, we even have $v(a+b) = min(v(a),v(b))$.

Because as you say $0 =\displaystyle \sum_{i=1}^\infty \dfrac{2^i}{i}$, this implies for any $n \in \mathbb N$: $$v_2(\displaystyle \sum_{i=1}^n\dfrac{2^i}{i}) = v_2(\displaystyle \sum_{t=n+1}^\infty \dfrac{2^t}{t}).$$

Now since further $v_2(\dfrac{2^t}{t}) = t-v_2(t)$, you only have to show that the term on the right is $$\ge \min_{t\ge n+1}(v_2(\dfrac{2^t}{t}))$$ which again follows from our ultrametric inequality and, to be rigorous, from the fact that the series converges i.e. that the individual summands go to zero i.e. their valuations go to $\infty$ for $t\to\infty$, i.e. the minimum exists and you can cut off the series anywhere after it without changing the series' valuation.

Torsten Schoeneberg
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