# An Infinite Product from Recursion

The following problem is from the first mathematics competition at the School of Mathematics and Statistics, The University of Sheffield. Here is a solution submitted by students Robert Nicolaides and Lawrence Beesely-Peck (Math Spectrum, v 47, 2014/2015, n 2, 71.)

The sequence $a_n$ is defined by $a_{1}=1$ and $a_{k+1}=(k+1)(a_{k}+1)$ for all $k\gt 1.$ Find $\displaystyle\prod_{k=1}^{\infty}\left(1+\frac{1}{a_k}\right).$

### Solution

Rewrite the recursion as

$\displaystyle 1+\frac{1}{a_k}=\frac{a_{k}+1}{a_{k}}=\frac{a_{k+1}}{(k+1)a_{k}}.$

Let $\displaystyle P(n)=\prod_{k=1}^{n}\left(1+\frac{1}{a_k}\right).$ Then we have

\displaystyle\begin{align} P(n) &= \frac{a_2}{2a_1}\frac{a_3}{3a_2}\frac{a_4}{4a_3}\cdots\frac{a_{n+1}}{(n+1)a_n}\\ &=\frac{a_{n+1}}{(n+1)!}\\ &=\frac{(n+1)(a_{n}+1)}{(n+1)!}\\ &=\frac{a_n}{n!}+\frac{1}{n!}\\ &=\frac{n(a_{n_1}+1)}{n!}+\frac{1}{n!}\\ &=\frac{a_{n-1}}{(n-1)!}+\frac{1}{(n-1)!}+\frac{1}{n!}\\ &\cdots\\ &=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\ldots+\frac{1}{n!}. \end{align}

So, as $\displaystyle\lim_{n\rightarrow\infty}P(n)=e.$ Hence, $\displaystyle\prod_{k+1}^{\infty}\left(1+\frac{1}{a_k}\right)=e.$