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Let a be a positive integer $a \geq 2$ and $S_{n}$ be a sum defined as follows $$S_{n}=\sum_{k=1}^{n}\frac{a!}{(a+k)(a+k-1)...(k+1)}$$ Find $\text{lim}_{n \rightarrow \infty} \, S_n$

I noticed that $(a+k)(a+k-1)...(k+1)=\frac{(a+k)!}{k!}$ and then $$\sum_{k=1}^{n}\frac{a!}{(a+k)(a+k-1)...(k+1)}=\sum_{k=1}^{n}\frac{a!k!}{(a+k)!}$$ I thought at the possibility of being a power sum, but I couldn't find it and this problem is for 11th graders. Another thing that I noticed is that $$\text{lim}_{n \rightarrow \infty} \,\frac{n!}{(a+n)(a+n-1)...(a+1)}=\text{lim}_{n \rightarrow \infty} \,\frac{a!n!}{(a+n)!}=0$$ I solved it by placing the expression between to expressions with their limits equal with $0$ $$0< \frac{n!}{(a+n)(a+n-1)...(a+1)}=\frac{1}{(1+a)(1+\frac{a}{2})...(1+\frac{a}{n})}< \frac{1}{1+a+\frac{a}{2}+...+\frac{a}{n}}$$ but this didn't help much. Also, I forcibly calculated the limit and it seems to be $\frac{1}{a-1}$. Any help would be appreciated.

omega123
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    Perhaps you will find some inspiration in https://math.stackexchange.com/questions/151441/calculate-sums-of-inverses-of-binomial-coefficients – Pedro Oct 06 '22 at 19:09
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    Hint: $$ \frac{{a!k!}}{{(a + k)!}} = \frac{{a!k!}}{{(a - 1)(a + k - 1)!}} - \frac{{a!(k + 1)!}}{{(a - 1)(a + k)!}} $$ (telescoping sum) – Gary Oct 06 '22 at 19:51
  • Thank you, i managed to solve it in both ways. I will post them when I have more time – omega123 Oct 06 '22 at 20:28

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