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Where $a_{1}=1$ and $a_{n+1}= \left ( n+1 \right )\left ( 1+a_{n} \right )$, $\forall \left ( n\epsilon \mathbb{N} \right )$

I have found that for this infinite product to be convergent, the series $\sum \ln \left ( 1+\frac{1}{a_{n}} \right)$ must also be convergent, and that if it has a finite sum, let's name it s, the value of the infinite product is $ e^{s}$. I am not sure whether $a_{n}=n\cdot \left ( 1+a_{n-1} \right )$ also holds true.

The given series passes the n-th term test, ok but how do I apply another test to it?

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  • $\begingroup$ @Arian It's the first one, I was wondering if the second one could be derived from the first one. $\endgroup$ – Shemafied Nov 25 '14 at 20:31
  • $\begingroup$ If you divide by $(n+1)!$ your relation, you get $\displaystyle \frac{a_{n+1}}{(n+1)!}=\frac{a_n}{n!}+\frac{1}{n!}$. Hence you get $a_n=n!(\sum_{k=0}^{n-1}\frac{1}{k!})$. You have $a_n\sim e n!$, and this gives the convergence of the infinite product, but not its value. $\endgroup$ – Kelenner Nov 25 '14 at 20:49
  • $\begingroup$ @Shemafied: yes you can do that derivation. It stands for the same recurrence relation. $\endgroup$ – Arian Nov 25 '14 at 21:11
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Note that $$a_{n+1}=(n+1)(1+a_n)\Rightarrow \frac{a_{n+1}}{1+a_{n}}=n+1$$ Consider the finite product:

$$\mathcal{P}_n=\prod_{n=1}^{N}(1+\frac{1}{a_n})=\prod_{n=1}^{N}\frac{1+a_n}{a_n}=\prod_{n=1}^{N}\frac{a_{n+1}}{(n+1)a_n}=\prod_{n=1}^{N}\frac{1}{(n+1)}\cdot \prod_{n=1}^{N}\frac{a_{n+1}}{a_n}=\frac{1}{N!}\cdot\frac{a_N}{a_1}=\frac{a_N}{N!}$$ Now we need to find an expression for the general term $a_n$. Notice that $$a_{n+1}=(n+1)(1+a_n)=(n+1)+(n+1)n(1+a_{n-1})=(n+1)+(n+1)n+(n+1)n(n-1)(1+a_{n-2})=...=(n+1)+(n+1)n+(n+1)n(n-1)+..+(n+1)n(n-1)(n-2)...2\cdot(1+1)$$ In other words $$a_{n+1}=\frac{(n+1)!}{n!}+\frac{(n+1)!}{(n-1)!}+..+2\cdot\frac{(n+1)!}{1!}$$ Hence $$a_N=\frac{N!}{(N-1)!}+\frac{N!}{(N-2)!}+..+2\cdot\frac{N!}{1!}$$ for all $N\geq1$ so $$\mathcal{P}_N=\frac{a_N}{N!}=\frac{1}{(N-1)!}+\frac{1}{(N-2)!}+..+\frac{2}{1!}$$ One can see that $$1\leq \mathcal{P}_N<\frac{1}{(N-1)(N-2)}+\frac{1}{(N-2)(N-3)}+..+2=3-\frac{1}{N-1}$$ then $$1\leq\lim_{N\to\infty}\mathcal{P}_N\leq3$$ Indeed the limit is $e$. Check it with the Taylor expansion of $e^x$ at $x=1$.

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