Why the sequence generated by $x+e^x$ is $(1,x,\frac{x^2}{2!},\frac{x^3}{3!},\frac{x^4}{4!}\ldots)$?

Question

I am trying to find the sequence generated by $x+e^x$. I have the sequence generated by the function $e^x$ which is $\displaystyle (1,x,\frac{x^2}{2!},\frac{x^3}{3!},\frac{x^4}{4!},\ldots)$, as for the sequence generated by $x$, I've seen one example in the book in which it's possible to obtain it doing the following:

$$\begin{eqnarray} {\frac{1}{1-x}}&=&{1+x+x^2+x^3+x^4+\ldots}\\ {}&&{}\\ {\left[ \frac{1}{1-x}\right]' }&=&{0+1+2x+3x^2+4x^3+\ldots}\\ {}&&{}\\ { \frac{1}{(1-x)^2} }&=&{0+1+2x+3x^2+4x^3+\ldots}\\ \end{eqnarray}$$

Now using: $f(x),g(x)$ generating functions of the sequences $(a_r),(b_r)$, we have: $Af(x)+Bg(x)$ Is the generating function to the sequence $(Aa_r+Bb_r)$.

I believe that this indicates that I should sum term by term:

$$\begin{eqnarray} {}&&{}\\ { \frac{1}{(1-x)^2}+e^x }&=&{[0+1]+[1+x]+[2x+\frac{x^2}{2!}]+[3x^2+\frac{x^3}{3!}]+[4x^3+\frac{x^4}{4!}]+\ldots}\\ \end{eqnarray}$$

But it doesn't add up to the desired sequence. What did I do wrong?

Answer 1

$x$ itself is a power series

$$0+x+0x^2+0x^3+\dots$$,

so

$$x+e^x=1+2x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\dots.$$

Answer 2

The sequence generated by $e^x$ is $$ (1,1,\frac{1}{2!},\frac{1}{3!},\frac{1}{4!},\ldots), $$ and the sequence for $x$ is just $$ (0,1,0,0,\ldots), $$ and thus the sequence generated by the sum, $x+e^x$, is just the sum of the sequences, $$ (1,2,\frac{1}{2!},\frac{1}{3!},\frac{1}{4!},\ldots). $$

I assume that you mean by "the sequence generated by $f(x)$", the sequence $(a_0,a_1,a_2,\ldots)$ such that we can expand $f(x)$ as $$ f(x)=\sum_{i=0}^{\infty} a_i x^i $$