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Let $p$ be a prime, $\mathbb C_p$ be the completion of a algebraic closure of $\mathbb Q_p$ and $(u_n)_{n\in\mathbb N}$ be a sequence of $\mathbb C_p$ converging towards $0$. Suppose that for all $n\in\mathbb N$, one has $1+u_n\ne0$. Can $\prod_{n\in\mathbb N}(1+u_n)$ be zero?

I know that is trivially true in $\mathbb C$, but I do not have any exemple (or counterexemple) in $\mathbb C_p$.

Thanks in advance.

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    $\begingroup$ Am I missing something? the product can be zero in $\mathbb C$: just take $u_n=-1/n$.. $\endgroup$ – Teri Jun 30 '16 at 5:59
  • $\begingroup$ My mistake. I had a problem with negation :) I rewrite the post. $\endgroup$ – joaopa Jun 30 '16 at 6:09
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    $\begingroup$ You're asking whether the sum of the $\log(1+u_n)$ (which is well defined for $n$ large enough) can diverge. But $\log(1+u_n)\to 0$ so the sum of them converges $p$-adically. Did I miss something? $\endgroup$ – Gro-Tsen Jun 30 '16 at 6:40
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    $\begingroup$ You say it's trivially true in C but in my mind the argument for C_p is even easier: can't you just mimic the argument you know in C? The product converges and you can even see what the norm of the product is, it's just the product of the norms of 1+u_n over the u_n for which |u_n|>=1. $\endgroup$ – znt Jun 30 '16 at 8:42
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If $n_0$ is such that $|u_n|<1$ for $n \geq n_0$, then $|1+u_n| = 1$ for $n \geq n_0$, so the norm of the infinite product is the norm of the product of the first $n_0$ terms and hence nonzero.

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