The generating function of $E_n = E_{n-1} + \frac12$, $E_1 = 1$

Question

I'm getting $\frac{2-x}{2(1-x)^2}$ and I can't figure what is wrong, $E(1) \neq 1$ in this case.

Answer 1

I'm unfamiliar with the method you describe in the comments. Here's how I would do it:

Let $f(x)$ be the generating function for $E_n$,

$$f(x)=\sum_{n\ge1}E_nx^n$$

Then for this recurrence, we have

$$\begin{align*} E_n&=E_{n-1}+\frac12\\[1ex] \sum_{n\ge2}E_nx^{n-1}&=\sum_{n\ge2}E_{n-1}x^{n-1}+\frac12\sum_{n\ge2}x^{n-1}\\[1ex] \frac1x\left(\sum_{n\ge1}E_nx^n-E_1x\right)&=\sum_{n\ge1}E_nx^n+\frac12\left(\sum_{n\ge0}x^n-1\right)\\[1ex] \frac{f(x)-x}x&=f(x)+\frac12\left(\frac1{1-x}-1\right)\\[1ex] f(x)&=\frac{2x-x^2}{2(1-x)^2}\\[1ex] \end{align*}$$

so that ultimately, your answer is just missing a factor of $x$.

Answer 2

with your function$$ f(x)= \frac{2-x}{2(1-x)^2}=$$

$$1+(3/2)x + (5/2)x^2+...=\sum _{n=0}^\infty \frac {(n+2)}{2} x^n$$

You have $E(0)=f(0)=1$

Thus if you want to get $$x+(3/2)x^2 + (5/2)x^3+...=\sum _{n=0}^\infty \frac {(n+2)}{2} x^{n+1}$$ You multiply your functin by $x$ to get $$g(x)= \frac{x(2-x)}{2(1-x)^2}$$