Prove that for any prime $p$ we have $|\sum_{n\ge 1} n^2(n+1)!|_p=2$

Question

This is a strange question and I don't know how to deal with this infinite series in p-adic evaluation. Any hint would be appreciated!

Answer 1

I think that your question should be:

Show that for any prime $p$, $|\sum_{n=1}^N n^2(n+1)!-2|_p\to 0$ where $|\cdot |_p$ is the $p$-adic order.

Since $n^2(n+1)!=(n-1)(n+2)!-(n-2)(n+1)!$, it follows that $$a_N:=\sum_{n=1}^Nn^2(n+1)!=\sum_{n=1}^N((n-1)(n+2)!-(n-2)(n+1)!)=(N-1)(N+2)!+2.$$ Hence $a_N-2=(N-1)(N+2)!$ converges $p$-adically to $0$, i. e. $a_N$ converges $p$-adically to $2$.