1. Jan 25, 2009

### caji

1. The problem statement, all variables and given/known data

Show, that in $$\mathbb{Q}_2$$ it holds

$$\sum_{i=1}^{\infty} \frac{2^i}{i} = 0$$

Hint: Consider $$\log_2(-1)$$

2. Relevant equations

Just for completeness:

$$\log(1+s) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{s^n}{n}$$

3. The attempt at a solution

Actually it's no homework but an exercise from a book. I found already, that

$$\lim_{i \to \infty} |2^i|_2 = 0$$

which seems to be correct as I found another thread in this forum called "p-adic convergence" which says the same. But that doesn't help me, because

$$\lim_{i \to \infty} \left|\frac{1}{i} \right|_2$$

does not converge.

Does anybody know, how to use this hint given in the exercise?

Ps.: Hmm, if one looks at $$\log(-1)$$, then in the upper formula, $$s=-2$$. That means $$\sum_{i=1}^{\infty} \frac{2^i}{i} = -\log(-1) \overset{?}{=} 0$$. Could that be correct?