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 $e^{1/a_1}$?
Best Answer
$\def\e{\mathrm{e}}$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$$ \prod_{k = 1}^n \left( 1 + \frac{1}{a_k} \right) = 1 + \frac{1}{a_1} \sum_{k = 0}^n \frac{1}{k!} \to 1 + \frac{\e}{a_1}. \quad n \to \infty $$