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Evaluate $\lim_{n\to\infty}(1+\frac{1}{a_1})(1+\frac{1}{a_2})\cdots(1+\frac{1}{a_n})$, where $a_1 = 1$, $a_n = n(1+a_{n-1})$

\begin{align*} &\lim_{n\to\infty} \left(1+\frac{1}{a_1}\right)\left(1+\frac{1}{a_2}\right)\cdots\left(1+\frac{1}{a_n}\right) \\ &= \lim_{n\to\infty} \frac{1}{a_1} \cdot \frac{1}{2} \cdot \frac{1}{3} \cdots \frac{1}{n} \cdot (a_n + 1) \\ &= \lim_{n\to\infty} \frac{a_n + 1}{n!}. \end{align*}

Then I'm stuck. How to proceed?

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    $\begingroup$ This is OEIS sequence A007526. The OEIS entry has much information including the limit you want to find. $\endgroup$ – Somos Jan 1 at 15:58
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Observe that $$a_n=n+na_{n-1}$$ $$a_n=n+n(n-1)+n(n-1)a_{n-2}$$ and so on, giving us eventually $$a_n=\sum_{k=1}^n\frac{n!}{(k-1)!}$$ Now we have $$\lim_{n\to\infty}\frac{1}{n!}(a_n+1)=\lim_{n\to\infty}\frac{a_n}{n!}$$ $$=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{(k-1)!}$$ $$=\sum_{k=0}^\infty\frac{1}{k!}=e$$

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  • $\begingroup$ Not $e+1$. The sum for $e$ includes the term with $0!$ as denominator. $\endgroup$ – Oscar Lanzi Jan 1 at 15:29
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    $\begingroup$ @OscarLanzi oops! Fixed it. $\endgroup$ – Ben W Jan 1 at 15:32
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Show by induction that

$a_1=1!/0!$

$a_2=2!/0!+2!/1!$

$a_3=3!/0!+3!/1!+3!/2!$

...

$\color{blue}{a_n=n!/0!+n!/1!+n!/2!+...+n!/(n-1)!}$

Thereby the limit is $1/0!+1/1!+2/2!+...1/n!+...=e$.

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From where you stuck you can proceed as \begin{align} \frac{a_n+1}{n!}&=\frac{a_n}{n!}+\frac{1}{n!}=\frac{1}{(n-1)!}\frac{a_n}{n}+\frac{1}{n!}=\frac{1}{(n-1)!}(a_{n-1}+1)+\frac{1}{n!}=\\ &=\frac{a_{n-1}}{(n-1)!}+\frac{1}{(n-1)!}+\frac{1}{n!}=\ldots= 1+\frac{1}{1!}+\ldots+\frac{1}{n!}. \end{align}

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