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$$ \mbox{Given the following sequence}:\quad a_{n + 1} = n + n\,a_{n}\quad\mbox{and}\quad a_{1} = 0. $$

  • How would one find the closed form of this $?$.
  • This looks vaguely like something involving factorials.
  • Finding the closed form is difficult, and I'm not sure how to start besides trial and error.

Any help would be appreciated.

Felix Marin
  • 89,464
Allen
  • 53

1 Answers1

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Using @Sil 23, letting $b_k=\frac{a_n}{k!}$,we can write $$b_{n+1}-b_n=\frac{1}{(n-1)!}$$ and carry out telescopic summation, we get $$b_2-b_1=1$$ $$b_3-b_2=\frac{1}{1!}$$ $$b_4-b_3=\frac{1}{2!}$$ $$............$$ $$b_n-b_{n-1}=\frac{1}{(n-2)!}$$ Adding all we get $$b_n=b_1+\sum_{k=0}^{n-2}\frac{1}{k!}$$ $$\implies b_n=e\frac{\Gamma(n-1,1)}{\Gamma(n-1)}\implies a_n=e \frac{\Gamma(n) \Gamma(n-1,1)}{\Gamma(n-1)}=(n-1)e~\Gamma(n-1,1).$$ Here, $\Gamma[n,x]$ are incomplete Gamma functions e.g. $\Gamma[1,1]=1/e$ and we get $$a_1=0, a_2=1, a_3=4, a_4=15,a_5=64,....$$

EDIT: Note that $\Gamma(n+1,1)=n\Gamma(n,1)+e^{-1}.$

Z Ahmed
  • 43,235