Recursion in limits

Calculus Level 4

If $a_1=1$ and $a_n=n(1+a_{n-1})$ for all $n \geq 2$ , then evaluate the following limit.

$\lim_{n \to \infty} \left[ \left(1+\frac{1}{a_1} \right) \left( 1+ \frac{1}{a_2} \right) \left( 1+ \frac{1}{a_3} \right) \cdots \left(1+\frac{1}{a_n} \right) \right]$

e^2

\frac{\pi}{2}

e

\pi

1 solution

Chew-Seong Cheong
Nov 11, 2019

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

$\begin{aligned} p_n & = \left(1+\frac 1{a_1} \right) \left(1+\frac 1{a_2} \right) \left(1+\frac 1{a_3} \right) \cdots \left(1+\frac 1{a_n} \right) \\ & = \frac {a_1+1}{a_1} \times \frac {a_2+1}{a_2} \times \frac {a_3+1}{a_3} \times \cdots \times \frac {a_n+1}{a_n} \\ & = \frac {\cancel{a_1+1}}{a_1} \times \frac {\cancel{a_2+1}}{2\cancel{(a_1+1)}} \times \frac {\cancel{a_3+1}}{3\cancel{(a_2+1)}} \times \cdots \times \frac {a_n+1}{n\cancel{(a_{n-1}+1)}} \\ & = \frac {a_n + 1}{n!a_1} = \frac {a_n+1}{n!} = \frac {n(a_{n-1}+1)+1}{n!} \\ & = \frac {a_{n-1}+1}{(n-1)!} + \frac 1{n!} = p_{n-1} + \frac 1{n!} \\ & = 1 + \frac 11 + \frac 1{2!} + \frac 1{3!} + \cdots + \frac 1{n!} \end{aligned}$

Therefore, $\displaystyle \lim_{n \to \infty} p_n = \boxed e$ .

Recursion in limits

1 solution

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