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Motivated by this question, I find the following recurrence relation for the exponential function (source, formula 25): \begin{align} a_n&=n(a_{n-1}+1)& 1+e/a_1=\prod_{n=1}^\infty(1+a_n^{-1}) \end{align} I'm looking for a proof of the formula on the right.

The explicit formula for $a_n$ is given by: \begin{align} \frac{a_n}{n!} &=\frac{a_{n-1}}{(n-1)!}+\frac 1{(n-1)!}\\ &=a_1+\sum_{k=0}^{n-1}\frac 1{k!}\\ &\to a_1+e \end{align} hence $a_n\sim (a_1+e)n!$ as $n\to\infty$. Any idea for to prove the convergence of the infinite product to $e^{1/a_1}$?

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$\def\e{\mathrm{e}}$For any $n \in \mathbb{N}_+$,$$ \prod_{k = 1}^n \left( 1 + \frac{1}{a_k} \right) = \prod_{k = 1}^n \frac{a_k + 1}{a_k} = \prod_{k = 1}^n \frac{a_{k + 1}}{(k + 1) a_k} = \frac{1}{a_1} \frac{a_{n + 1}}{(n + 1)!}, $$ and as is already proved,$$ \frac{a_{n + 1}}{(n + 1)!} = a_1 + \sum_{k = 0}^n \frac{1}{k!}, $$ thus$$ \prod_{k = 1}^n \left( 1 + \frac{1}{a_k} \right) = 1 + \frac{1}{a_1} \sum_{k = 0}^n \frac{1}{k!} \to 1 + \frac{\e}{a_1}. \quad n \to \infty $$

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  • $\begingroup$ So the infinite product does not converge to $e^{1/a_1}$, or did I miss something? $\endgroup$ – awllower Jan 31 '18 at 14:21
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    $\begingroup$ @awllower You're right. $\endgroup$ – Saad Jan 31 '18 at 14:24
  • $\begingroup$ Is possible to modify the recurrence in order to obtain $e^{1/a_1} $ as limit? $\endgroup$ – Fabio Lucchini Feb 2 '18 at 7:18

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