Time complexity of $T(n) = nT(n-1) + n^2$ This time complexity comes from $N$-queen problem. There are many threads talking about the time complexity to be $O(n!)$ but I cannot figure out how we get it. Here is the progress I have so far:

$$
\begin{align}
T(n) &amp; = n T(n-1) + n^2\\
&amp; = n(n-1)T(n-2) + n(n-1)^2 + n^2\\
&amp; \vdots\\
&amp; = n! + n! + n(n-1) \dots 2^2 + n(n-1) \dots 4*3^2 + \dots + n(n-1)^2 + n^2
\end{align}
$$

RHS and LHS divided by $n!$

$$
    \frac{T(n)}{n!} = 2 + \frac{2}{1!} + \frac{3}{2!} + \dots + \frac{n-2}{(n-3)!} + \frac{n-1}{(n-2)!} + \frac{n}{(n-1)!}
$$

From comments in GeeksForGeeks the RHS converges to a constant. Can anyone please explain why the RHS converges to a constant?
We can compute the limit explicitly:
$$
1 + \sum_{n=0}^\infty \frac{n+1}{n!} = 1 + \sum_{n=0}^\infty \frac{1}{n!} + \sum_{n=1}^\infty \frac{1}{(n-1)!} = 2e + 1.
$$