<s>[INST] Question about the exponential generating function for $T_n=T_{n-1}+(n-1)\cdot T_{n-2}$ So I came across this step when finding an expontential generating function using this recurrence relation,
$$
T_{n}=T_{n-1}+(n-1) \cdot T_{n-2}
$$
Since $T_{0}=T_{1}=1$
For $n=0$
$$
\sum_{n \geq 0} \frac{T_{0} x^{0}}{0 !}=1
$$
For $n=1$
$$
\sum_{n \geq 1} \frac{T_{1} x^{1}}{1 !}=x
$$
Therefore,
$$
f(x)=\sum_{n \geq 0} \frac{T_{n} x^{n}}{n !}=1+x+\sum_{n \geq 2} \frac{T_{n} x^{n}}{n !}
$$
From the recurrence relation, $T_{n}=T_{n-1}+(n-1) \cdot T_{n-2}$. Hence,
$$
\Rightarrow 1+x+\sum_{n \geq 2} \frac{T_{n-1}+(n-1) T_{n-2}}{n !} x^{n}
$$
Distributing the terms using the commutative law of addition, we get,
$$
\begin{gathered}
\therefore 1+x+\sum_{n \geq 2} \frac{T_{n-1}}{n !} x^{n}+\sum_{n \geq 2} \frac{(n-1) T_{n-2}}{n !} x^{n} \\
\Rightarrow 1+\sum_{n \geq 1} \frac{T_{n-1}}{n !} x^{n}+\sum_{n \geq 2} \frac{(n-1) T_{n-2}}{n !} x^{n}
\end{gathered}
$$
Where did the ' $x$ ' go in the last step above? And how did the lower limit change from $n \geq 2$ to $n \geq 1$ ?
Could anyone help explain? Any help would be highly appreciated.
 [/INST] Just compare the two lines, focusing on
$$\sum_{n\geq 1} \frac{T_{n-1}}{n!}x^n = \color{red}{\underbrace{\frac{T_{1-1}}{1!}x^1}_{n=1}} + \color{blue}{\underbrace{\frac{T_{2-1}}{2!}x^2}_{n=2} + \underbrace{\frac{T_{3-1}}{3!}x^3}_{n=3} + \cdots}$$
$$x + \sum_{n\geq 2} \frac{T_{n-1}}{n!}x^n = \color{red}{x} + \color{blue}{\underbrace{\frac{T_{2-1}}{2!}x^2}_{n=2} + \underbrace{\frac{T_{3-1}}{3!}x^3}_{n=3} + \cdots}$$
so it all comes down to $$\frac{T_{1-1}}{1!}x^1 \stackrel{?}{=} x$$ and this is true since $T_0 = 1$.
</s>