## another question on indices on generating functions

Use the recurrence $D_n = n D_{n-1} + (-1)^n, n\geq 2$ to find the exponential generating function for $D_n$.

Here is my solution: $\sum_{n \geq 2} D_n \frac{x^n}{n!} = \sum_{n \geq 2} n D_{n-1} \frac{x^n}{n!} + \sum_{n \geq 2} (-1)^n \frac{x^n}{n!} \\
\implies & G_D(x) = x \sum_{n \geq 2} D_{n-1} \frac{x^{n-1}}{(n-1)!} + (e^{-x} -e^{-1}-1)\\
\implies & G_D(x) = x \sum_{n \geq 1} D_{n} \frac{x^{n}}{n!} + (e^{-x} -e^{-1}-1) \\
\implies & G_D(x) = x (D_{1} \frac{x^{1}}{1!}+\sum_{n \geq 2} D_{n} \frac{x^{n}}{n!}) + (e^{-x} -e^{-1}-1) \\
\implies & G_D(x) = x G_D(x) + (e^{-x} -e^{-1}-1) \\
\implies & G_D(x) = \frac{e^{-x} -e^{-1}-1}{1-x}
$

But when I looked at the solution, the $-e^{-1}-1$ was unnecessary.
This happened because I began the summation from 2.
I did this because (i) the problem stated $n\geq 2$, and (ii) $D_{-1}$ wouldn't make sense.
Also, if the problem asked me to find "the generating function for $n\geq 2$", does that mean
$G_D(x)=\sum_{n \geq 2} D_n \frac{x^n}{n!}$
$G_D(x)=\sum_{n \geq 0} D_n \frac{x^n}{n!}$