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To what function does this Power Series converges to $P(x)=\sum_{j=0}^{j=\infty}(j+1)(j+2)x^j$

I know its nothing but $P(x)=\sum_{j=0}^{j=\infty}\frac{(j+2)!}{j!}x^j$

So what is the function it converges to and How?

  • 3
    Compute $\frac{d^{2}}{dx^{2}}\sum_{j\geq 0}x^{j+2}$ a recall the geometric serie – 2017-01-27
  • 0
    But what does $\sum_{j\geq 0}x^{j+2}$ converge to? I don't recall a proper function – 2017-01-27
  • 1
    Hey, I think I got it $\sum_{ j\geq 0}x^{j+2}$ converges to $$\frac{x^2}{1-x}$$ Hence double integral of it will give the solution. Correct? – 2017-01-27
  • 0
    Double derivative... – 2017-01-27

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It is know that $\sum_{n\geq 0}x^n=\frac{1}{1-x}$. Then, we have in one hand $$ \frac{d^{2}}{dx^{2}}\sum_{n\geq 0}x^{n+2}=\frac{d^{2}}{dx^{2}}\left (x^2\sum_{n\geq 0}x^{n}\right )=\frac{d^{2}}{dx^{2}}\left ( x^2\frac{1}{1-x}\right )$$ and in the other hand $$ \frac{d^{2}}{dx^{2}}\sum_{n\geq 0}x^{n+2}=\sum_{n\geq 0}\frac{d^{2}}{dx^{2}}x^{n+2}=\sum_{n\geq 0}(n+2)(n+1)x^{n} $$