I have this recurrence relation:

$$\begin{equation} \begin{cases} a_n = 2a_{n-1} + n & (n\geq1)\\ a_0 = 1 \end{cases} \end{equation}$$


$$f(x) = \sum_{n=0}^\infty a_nx^n$$

I solved in this way:

$$\sum_{n=1}^\infty a_nx^n = 2\sum_{n=1}^\infty a_{n-1}x^n + \sum_{n=1}^\infty nx^n$$ $$f(x)-1 = \frac{2}{x}\big(f(x)-1\big) + \frac{1}{1-x}-1$$ $$f(x)-1 = \frac{2}{x}\big(f(x)-1\big) + \frac{1}{(1-x)^2}-1$$ $$f(x)\big(1-\frac{2}{x}\big) = \frac{-2(1-x)^2+x}{x(1-x)^2}$$

$$f(x) = \frac{-2x^2+5x-2}{(1-x)^2(x-2)}$$

and then I get:

$$f(x) = \frac{A}{1-x}+\frac{B}{(1-x)^2}+\frac{C}{x-2}$$

but the result I get is:

$$\begin{equation} \begin{cases} A = 2\\ B=-1\\C=0 \end{cases} \end{equation}$$

Don't think that $c=0$ is possible, where is the mistake?


I think that mistake is at $\sum_{n=1}^\infty nx^n$:

\begin{eqnarray}\sum_{n=1}^\infty nx^n &=& x\sum_{n=1}^\infty nx^{n-1}\\ &=& x\big(\sum_{n=0}^\infty x^{n}\big)'\\ &=& x\big({1\over 1-x}\big)'\\ &=& {x\over (1-x)^2} \end{eqnarray}

  • $\begingroup$ Yeah, that's the mistake. now I get $\frac{x^3-3x^2+x-2}{x^3-4x^2+5x-2}$ how should I procede? $\endgroup$ – Christian Giupponi Jan 27 '18 at 15:49
  • $\begingroup$ $$\frac{x^3-3x^2+x-2}{x^3-4x^2+5x-2} = 1+\frac{x^2-4x}{x^3-4x^2+5x-2}$$ $\endgroup$ – Aqua Feb 1 '18 at 13:48
  • $\begingroup$ Now $$ \frac{x^2-4x}{x^3-4x^2+5x-2} = {a\over x-1}+ {b\over (x-1)^2}+{c\over x-2}$$ $\endgroup$ – Aqua Feb 1 '18 at 13:49
  • $\begingroup$ sorry @christian I have noticed an error in my excercise. the second sum should be $2x$ instead of $\frac{2}{x}$? $\endgroup$ – Christian Giupponi Feb 5 '18 at 10:20

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