I am working on the convergence of the following series in $Q_p$: $a_n$ = $1/n$ and $a_n$ = $(1+p)^{p^n}$
For the first one, my solution is as follows: $$\lim_{n\to\infty} | (\frac{1}{n+1})-(\frac{1}{n})|_p = \lim_{n\to\infty} | \frac{1}{n(n+1)}|_p = \lim_{n\to\infty}p^{-v_p(1/n)} p^{-v_p(1/(n+1))}$$ and as $n \to \infty$ $1/n \to 0$ also $1/(n+1) \to 0$ then $v_p(0) = \infty$ thus answer is $\infty$.
For the second one, I considered $$\lim_{n\to\infty} |(1+p)^{p^n}[((1+p)^{p^n})^{p-1} -1]|_p = \lim_{n\to\infty} p^{-v_p((1+p)^{p^n})}p^{-v_p(((1+p)^{p^n})^{p-1} -1)}$$
now using Fermat's little theorem we know that p will divide this expresssion $(((1+p)^{p^n})^{p-1} -1)$ but p will not divide $(1+p)^{p^n}$ becuase p does not divide $(1+p)$. But as $n \to \infty$ then $(1+p)^{p^n}$ also goes to infinity so p divides this expression infinitely many times thus the answer is $0$.
Could anyone please tell me if the solution is correct. Thank you.
What is correct is that for each $n$, $$ | (\frac{1}{n+1})-(\frac{1}{n})|_p = | \frac{1}{n(n+1)}|_p = p^{-v_p(1/n)} p^{-v_p(1/(n+1))} = p^{-v_p(1/n)-v_p(1/(n+1))}$$
It is also true that $1/n \to 0$ for $n \to \infty$ in the usual archimedean value; but, unfortunately, that is irrelevant to our problem. Think of $v_p$ as a function. Surely you know some functions $f$ such that $f(1/n) \not \to f(0)$ for $n \to \infty$? Right, that happens when $f$ is not continuous at $0$. And $v_p$ is such a function.
(I know that there are other problems here, like, where is it even defined, and continuous with respect to what metric, but I don't want to digress here.)
So what we actually want to know is whether that thing in the exponent, $$-v_p(1/n)-v_p(1/(n+1))$$ has any predictable behaviour for big $n$ (optimally, we would like it to go to $-\infty$, because then "$p$ to that power" would go to $0$, and our sequence would be a Cauchy sequence). Unfortunately, it does not. For example, if $n$ is a power of $p$, say $n= p^k$, then $$-v_p(1/n)-v_p(1/(n+1)) = k -0 = k$$ So if we look at indices $n = p, p^2, p^3, ...$, , the absolute values of the differences $$ | (\frac{1}{n+1})-(\frac{1}{n})|_p = | \frac{1}{n(n+1)}|_p = p^{-v_p(1/n)-v_p(1/(n+1))} = p^{k}$$ actually become bigger and bigger. That is, the subsequence $a_p, a_{p+1}, a_{p^2}, a_{p^2+1}, a_{p^3}, a_{p^3+1}, ...$ diverges, and hence the whole sequence diverges.
As for the second sequence, again the equation that you write would be correct at least without the "lim" in front of it, but the argument of "goes to infinity" and "p divides this expression infinitely many times" is wrong. I mean, I hope you would not say that the sequence $a_n =n$ converges $p$-adically for the same reason! What you actually want to find here is a good estimate of $v_p((1+p)^{p^n(p-1)}-1)$, for which you probably need to know the $p$-adic value of binomial coefficients. My guess, by the way, is that that value ($v_p((1+p)^{p^n(p-1)}-1)$) goes to infinity (but this is the hard part to show, with estimates of binomial coefficients), whereas $v_p((1+p)^{p^n})$ is constant $\equiv 0$ (this should be really easy to see), so indeed those differences go to zero, so the sequence is a Cauchy sequence and will converge. My further guess is that the limit is actually $1$. And one could prove that directly by showing that $$|a_n-1|_p =|(1+p)^{p^n}-1|_p$$ goes to $0$, or equivalently, $$v_p((1+p)^{p^n}-1)$$ goes to infinity for $n \to \infty$, with the same (harder) argument involving binomial coefficients as before. Namely, you have to get an estimate for $$\min \{v_p\left(\binom{p^n}{k}p^k\right): k=1, ..., p^n\} $$ and show that this grows without bounds for $n \to \infty$.