To show that in the field $\mathbb{Q}_p$, where $p$ is a prime, it holds that: $$-1=\sum_{0}^{\infty} (p-1)p^i$$ I did the following:
It suffices to show that:
$\left|\sum_0^N (p-1)p^i+1 \right|_p \to 0, \text{ when } n \to +\infty$
For each non-negative integer $N$,
$\sum_{0}^N (p-1)p^i=\sum_{0}^N (p^{i+1}-p^i)=p^{N+1}-1$
Thus: $\left| \sum_{0}^{N} (p-1)p^i-(-1)\right|_p=|p^{N+1}-1+1|_p=|p^{N+1}|_p=p^{-(N+1)} \to 0, \text{ when } N \to +\infty$
But the prof said that we cannot do it like that since we can't conclude from $|a_n|_p \to 0$ that $a_n \to 0$. How else could we do it?
First of all $\sum_{0}^{\infty} p^i = \frac{1}{1-p}$. Indeed, as $(1-p) \sum_{0}^{N} p^i = 1 - p^{N+1}$ and as $p^{N+1}\to 0$ in $\mathbf{Q}_p$ as $N\to +\infty$, we have $\sum_{0}^{\infty} p^i = \frac{1}{1-p}$.
Now : $\sum_{0}^{\infty} (p-1)p^i = (p-1) \sum_{0}^{\infty} p^i = (p-1) \frac{1}{1-p} = -1$ in $\mathbf{Q}_p$