How to find $$\lim\limits_{n\to\infty} \left(1+\frac{ 1 }{ a_{ 1 } } \right) \left( 1+\frac { 1 }{ a_{ 2} } \right)\cdots\left( 1+\frac { 1 }{ a_{ n } } \right) $$ where $$a_1=1$$ $$a_n=n(1+a_{n-1})$$ for all $n \geq 2$?
Asked
Active
Viewed 84 times
3
-
2Write the general factor as $\frac{a_n + 1}{a_n}$ and look for cancellation. – Daniel Fischer Jun 24 '15 at 15:27
-
1Hint: Start with @DanielFischer 's comment and then substitute into the denominator of $\frac{a_n+1}{a_n}$. – Michael Burr Jun 24 '15 at 15:29
-
Please check if my edit is correct. – k170 Jun 24 '15 at 15:30
-
I'm getting like $(1/n!+(a_{n})/n!)$ – Jun 24 '15 at 15:32
-
Okay, then you obviously need to find $\lim\limits_{n\to\infty} \dfrac{a_n}{n!}$ or show that limit doesn't exist. – Daniel Fischer Jun 24 '15 at 15:35
-
1Wait a sec...is it e ? :-D @DanielFischer – Jun 24 '15 at 15:38
1 Answers
4
When looking for the limit of an infinite product (or a series), it is often helpful to look for cancellation/telescoping. If telescoping occurs, that can give you a much simpler expression for the partial products (partial sums), from which the limit may be easy to determine.
Here, rewriting the factors as
$$\frac{a_k + 1}{a_k}$$
and using the recurrence $a_k = k(1+a_{k-1})$ (for $k > 1$) gives us
$$\prod_{k=1}^n \biggl(1 + \frac{1}{a_k}\biggr) = \frac{1+a_1}{a_1}\prod_{k = 2}^n \frac{a_k+1}{k(a_{k-1}+1)} = \frac{a_n + 1}{n!}.$$
So we see that we want to find $\lim\limits_{n\to\infty} \dfrac{a_n}{n!}$. For those terms, the recurrence becomes
$$\frac{a_n}{n!} = \frac{1}{(n-1)!} + \frac{a_{n-1}}{(n-1)!},$$
which together with the initial value $\frac{a_1}{1!} = \frac{1}{0!}$ yields
$$\frac{a_n}{n!} = \sum_{k = 0}^{n-1} \frac{1}{k!},$$
from which the limit $e$ is immediately read off.
Daniel Fischer
- 206,697