- #1

caji

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## Homework Statement

Show, that in [tex]\mathbb{Q}_2[/tex] it holds

[tex]\sum_{i=1}^{\infty} \frac{2^i}{i} = 0[/tex]

Hint: Consider [tex]\log_2(-1)[/tex]

## Homework Equations

Just for completeness:

[tex]\log(1+s) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{s^n}{n}[/tex]

## The Attempt at a Solution

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

[tex]\lim_{i \to \infty} |2^i|_2 = 0[/tex]

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

[tex]\lim_{i \to \infty} \left|\frac{1}{i} \right|_2[/tex]

does not converge.

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

Thanks in advance.

caji

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

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