W. H. Schikhof in his book on Ultrametric calculus suggests to solve the following problem:
Exercise 23.J (van Hamme) Use the ideas of the previous excercise to show that in $\mathbb Q_p$ ($p \neq 2$): $$S_p := \sum_{n=1}^\infty \frac{n^2 (n+1)!}{4^n+1} = -1.$$
The previous excercise simply states that in every $\mathbb Q_p$ we have $\sum_{n \ge 0} n \cdot n! = -1$. The series clearly converges at least in $\mathbb Q_3$, where $|4^n + 1|_3 = 1$ regardless of $n$ and $|(n+1)!|_3$ tends to zero, and we can invoke the famous ultrametric theorem concerning convergence of series (the necessary convergence condition is sufficient too).
However investigating the series $S_3$ in Pari-GP reveals that it doesn't really gets closer to $-1$, as $$1 + S_3 = 2 + 3 + 2\cdot 3^3 + 3^4 + 3^5 + 3^7 + 2 \cdot 3^{10} + \ldots $$
In that case I have to ask what is the proper limit of $S_p$ and where can I find the original paper written by van Hamme where he implicitly derives the value of $S_p$?
The mistake was indeed very minor. I am almost sure that after correction the exercise shall state
$$ \sum_{n=1}^\infty \frac{n^2(n+1)!}{4^{n+1}} = -4. $$
For a proof see this answer. The series does not converge in $\mathbb Q_2$ because the necessary condition is not fulfilled: if we take $n = 2k +1$ then
$$ \begin{align} v_2 \left(\frac{n^2 (n+1)!}{2^{2n + 2}}\right) & \le 2v_2(n) (n+1) - (2n + 2)\\ & = 0, \end{align} $$ so infinitely often the $n$-th summand has $2$-adic norm equal to $1$ (or something larger).