$2\cdot(n-2)/n!\; ?$
Convergence to $?$. It's part of a larger problem but the rest are all null sequences. The serie is : $u(n+1) = 1 + 1/n\cdot u(n),\qquad (u(n) \text{ is the }n\text{-th} \text{ term}), \qquad u(1) = 2.$
I managed to redefine the serie to a formula in $n$, being
$1/n + 1/n\cdot(n-1) + 2\cdot(n-2)/n! + 1/n\cdot(n-1)\cdot(n-2).$ I assume apart from the problem with $n!$ that all others are null-sequences.
By induction I proved $u(n) \le 3$.
So, if it converges to $2$ or $3$, some of the terms in the formula must be $\ge 1$. I think thus, that $2\cdot(n-2)/n!$ cannot be a null-sequence.