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For a prime $p$ , I need to show that $$\sum_{n=0}^{\infty}p^n$$ converges in $\mathbb{Z_p}$ (p-adic integers) .

Now since $$\sum_{n=0}^{\infty}p^n = \frac{1}{1-p} \in \mathbb{Z_p} $$ so I suppose one way to show the convergence is to show that this is a Cauchy sequence. But I am not able to do so.

Any suggestions!

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    What did you tried? – 2017-02-05
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    Do you want to show $\sum_{k=0}^{n}p^k$ is a Cauchy sequence ? – 2017-02-05
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    Yes. I was thinking of using $d(x,y) = e^{-v_p(x-y)}$ metric as $| \space |$ norm but not sure if this is the right approach. – 2017-02-05
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    Hint :$$a_n=\sum_{k=0}^{n}p^k \\ \forall m,n>N_0,m>n :|a_m-a_n|<\epsilon\\ |\sum_{k=0}^{m}p^k-\sum_{k=0}^{n}p^k|<\epsilon$$ $$|\sum_{k=0}^{m}p^k-\sum_{k=0}^{n}p^k|=\\|(1+p+p^2+p^3+...p^m)-(1+p+p^2+...+p^n))|=\\|p^{n+1}+p^{n+2}+...+p^m|<|1+p+p^2+...+p^m+...|=|\dfrac{1}{1-p}|=\dfrac{1}{1-p}$$ – 2017-02-05
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    @Khosrotash: The p-adic numbers do not follow the metric of the value. Also, the usual order of numbers is not compatible with the p-adic metric. Further consider that as a prime number $p>1$. – 2017-02-05

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The valuation of the distance of the $n$-th partial to the limit is $n+1$, as the difference is $\frac{p^{n+1}}{p-1}$. This already proves convergence in the $p$-adic metric.

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    I think you need to point out to OP that $\sum a_n$ converges if and only if $|a_n| \rightarrow 0$ with the $p$-adic norm – 2017-02-05
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    @MattB : This could be a different answer as it is more general and also applies if the limit is not known explicitly. I was using the ordinary definition of convergence in metric spaces. – 2017-02-05