Determine whether the below series converges or diverges. If it converges, find its sum.
\[
\sum_{n=1}^{\infty} \frac{n+1}{n !}
\]
a. Diverges
b. Converges to 2
c. Converges to $2 \mathrm{e}$
d. Converges to $2 e-1$

Step 1 ：Let's denote the sum of the series as \(S_m = \sum_{n = 1}^m \frac{n + 1}{n!}\).

Step 2 ：We compute the first few sums \(S_m\):

Step 3 ：\[\begin{array}{c|c} m & S_m \\ \hline 1 & 2 \\ 2 & 3 \\ 3 & \frac{8}{3} \\ 4 & \frac{17}{6} \\ 5 & \frac{71}{24} \end{array}\]

Step 4 ：We note that the fractions seem to be approaching 2, so we also compute \(2 - S_m\):

Step 5 ：\[\begin{array}{c|c|c} m & S_m & 2 - S_m \\ \hline 1 & 2 & 0 \\ 2 & 3 & -1 \\ 3 & \frac{8}{3} & -\frac{2}{3} \\ 4 & \frac{17}{6} & -\frac{5}{6} \\ 5 & \frac{71}{24} & -\frac{23}{24} \end{array}\]

Step 6 ：We can relate the fractions \(2 - S_m\) to factorials in the following way:

Step 7 ：\[-1 = -1, \ -\frac{2}{3} = -\frac{2}{3!}, \ -\frac{5}{6} = -\frac{5}{4!}, \ -\frac{23}{24} = -\frac{23}{5!}.\]

Step 8 ：Thus, we conjecture that \[S_m = 2 + \frac{m}{(m + 1)!}.\]

Step 9 ：So, let \[T_n = 2 + \frac{n}{(n + 1)!}.\]

Step 10 ：Then \[T_n - T_{n - 1} = \left( 2 + \frac{n}{(n + 1)!} \right) - \left( 2 + \frac{n - 1}{n!} \right) = \frac{n}{n!} - \frac{n - 1}{n!} = \frac{1}{n!},\] which is exactly what we are summing.

Step 11 ：From the identity \[\frac{1}{n!} = \frac{n}{(n + 1)!} - \frac{n - 1}{n!},\] we have that

Step 12 ：\[\sum_{n = 1}^\infty \frac{n + 1}{n!} = \left( 2 - 1 \right) + \left( 3 - 2 \right) + \left( \frac{8}{3} - 3 \right) + \dotsb = \boxed{2}.\]