recursion equation generator function

Question

\begin{eqnarray*} a_n=a_{n-1}+n, a_1=1 \end{eqnarray*} let´s $A(x)=\sum_{n=1}^{\infty}a_{n}x^n=a_1{x}+\sum_{n=2}^{\infty}a_{n}x^n=x+\sum_{n=2}^{\infty}a_{n}x^n$ $$A(x)=x+\sum_{n=2}^{\infty}\left(a_{n-1}+n\right)x^n=x+\sum_{n=2}^{\infty}a_{n-1}x^n+\sum_{n=2}^{\infty}nx^n$$ $$A(x)=x+\sum_{n=1}^{\infty}a_{n}x^{n+1}+\sum_{n=1}^{\infty}nx^n-x\Rightarrow$$ $$\Rightarrow A(x)=x\sum_{n=1}^{\infty}a_{n}x^{n}+\sum_{n=1}^{\infty}nx^n$$ we know that $\sum_{n=1}^{\infty}nx^n=x+2x^2+3x^3+4x^4+\ldots$ and $$\sum_{n=1}^{\infty}nx^n=\sum_{n=1}^{\infty}x\left(\frac{d}{dx}\right)x^n=x\left(\frac{d}{dx}\right)\sum_{n=1}^{\infty}x^n=x\left(\frac{d}{dx}\right)\frac{1}{1-x}=\frac{x}{\left(1-x\right)^2}$$ thus $$A(x)=xA(x)+\frac{x}{\left(1-x\right)^2}\Rightarrow \left(1-x\right)A(x)=\frac{x}{\left(1-x\right)^2}$$ $$A(x)=\frac{x}{\left(1-x\right)^3}$$ $$\frac{x}{\left(1-x\right)^3}=\frac{B}{\left(1-x\right)^3}+\frac{C}{\left(1-x\right)^2}+\frac{D}{\left(1-x\right)}$$ $$x=B+C(1-x)+D(1-x)^2\Rightarrow x=(B+C+D)-(C+2D)x+Dx^2$$ \begin{eqnarray*} \begin{cases} B+C+D=0\\ C+2D=-1\\ D=0 \end{cases} \Rightarrow \begin{cases} B=1\\ C=-1\\ D=0 \end{cases} \end{eqnarray*} $$A(x)=\frac{1}{\left(1-x\right)^3}-\frac{1}{\left(1-x\right)^2}$$ Somebody know what´s the wrong?

Answer 1

If you define $\left(a_n\right)_{n \in \mathbb{N}^{*}}$ as you did, then we have $$ a_n=\sum_{k=1}^{n}k=n\frac{n+1}{2} $$ Hence $$ A\left(x\right)=\frac{1}{2}\sum_{n=1}^{+\infty}n\left(n+1\right)x^n=\frac{1}{2}\sum_{n=1}^{+\infty}n^2x^n+\frac{1}{2}\sum_{n=1}^{+\infty}nx^n $$ As you wrote we have $$ \sum_{n=1}^{+\infty}n^2x^n=-\frac{x\left(x+1\right)}{(x-1)^3} \text{ and }\sum_{n=1}^{+\infty}nx^n=\frac{x}{(x-1)^2}=\frac{x\left(x-1\right)}{(x-1)^3} $$ Finally

$$ A(x)=\frac{-x\left(x+1\right)+x\left(x-1\right)}{2(x-1)^3}=-\frac{2x}{2(x-1)^3}=-\frac{x}{(x-1)^3}=\frac{x}{(1-x)^3} $$

Why would it be wrong ?

Answer 2

There's nothing wrong, you just have to note that you have two (generalized) binomial terms, so

\begin{align} A(x)&= (1-x)^{-3}-(1-x)^{-2} = \sum_{n=0}^{\infty} \binom{-3}{n} (-x)^n - \sum_{n=0}^{\infty} \binom{-2}{n} (-x)^n\\ &= \sum_{n=0}^{\infty} \underbrace{\left[\binom{-3}{n} - \binom{-2}{n}\right](-1)^n}_{a_n} \cdot x^n \end{align}

We have for $n\geq 1$ $$\binom{-3}{n} - \binom{-2}{n} \overset{1}{=} - \binom{-3}{n-1} \overset{2}{=} - (-1)^{n-1} \binom{n+1}{n-1} \overset{3}{=} (-1)^n\binom{n+1}{2} $$

Hence $$a_n = \left[(-1)^n\binom{n+1}{2}\right]\cdot(-1)^n = \binom{n+1}{2} = \frac{n(n+1)}{2}$$

where we used the well-known binomial coefficient facts

1. $\binom{r}{k} + \binom{r}{k-1} = \binom{r+1}{k}$
2. $\binom{-r}{k} = (-1)^k \binom{r+k-1}{k}$
3. $\binom{r}{k} = \binom{r}{r-k}$

Answer 3

Given that

$\frac{1}{\left(1-x\right)^2}=\frac{d}{dx}\left(\frac{1}{1-x}\right)=\frac{d}{dx}\left(\sum_{n=0}^{\infty}x^n\right)=\sum_{n=1}^{\infty}nx^{n-1}=\sum_{n=0}^{\infty}\left(n+1\right)x^n$

and

$\frac{1}{\left(1-x\right)^3}=\frac{1}{2}\frac{d^2}{dx^2}\left(\frac{1}{1-x}\right)=\frac{1}{2}\frac{d}{dx}\left(\frac{d}{dx}\left(\frac{1}{1-x}\right)\right)=\frac{1}{2}\frac{d}{dx}\left(\sum_{n=0}^{\infty}\left(n+1\right)x^n\right)=\frac{1}{2}\sum_{n=1}^{\infty}n\left(n+1\right)x^{n-1}$

$\dfrac{1}{\left(1-x\right)^3}=\sum_{n=0}^{\infty}\frac{\left(n+1\right)\left(n+2\right)}{2}x^n$

therefore

$$A(x)=\frac{1}{\left(1-x\right)^3}-\frac{1}{\left(1-x\right)^2}=\sum_{n=0}^{\infty}\frac{\left(n+1\right)\left(n+2\right)}{2}x^n-\sum_{n=0}^{\infty}\left(n+1\right)x^n$$ $$A\left(x\right)=\sum_{n=0}^{\infty}\left[\frac{\left(n+1\right)\left(n+2\right)}{2}-\frac{2\left(n+1\right)}{2}\right] x^n$$ $$A\left(x\right)=\sum_{n=0}^{\infty}\frac{\left(n+1\right)\left(n+2-2\right)}{2}x^n$$ $$A\left(x\right)=\sum_{n=0}^{\infty}\frac{n\left(n+1\right)}{2}x^n$$

so, $a_n=\dfrac{n\left(n+1\right)}{2}$.