Limit of a sequence $a_1=1;a_{n+1}=(n+1)(1+a_n)$

Question

Let a sequence be $a_1=1;a_{n+1}=(n+1)(1+a_n)$ If $P_n=\prod_1^n(1+a_i^{-1})$ then $$\lim_{n\to\infty}P_n $$is ?

I did: $$P_n=\prod_1^n\frac{(1+a_i)}{a_i} =\frac{a_{n+1}}{a_1}\prod1^{n-1}\frac{(1+a_i)}{a_{i+1}} =\frac{a_{n+1}}{a_1}\prod_1^{n-1}\frac1{i+1}=\frac{a_{n+1}}{n!}$$ Now how do I find the limit?

score 6 · Accepted Answer · answered Jun 20 '14 at 18:15

6

$a_n = n + na_{n-1}$. This is $a_n = n + n(n-1) + n(n-1)(n-2) + \ldots + n!$. In turn, $\frac {a_n}{n!} = \frac{1}{(n-1)!} + \frac{1}{(n-2)!} + \frac{1}{(n-3)!} + \ldots + 1$, which approaches the Taylor series for $e$.

answered Jun 20 '14 at 18:15

Karolis Juodelė

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25
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score -1 · Answer 2 · answered Jun 20 '14 at 18:26

-1

Let a sequence be $a_1=1;a_{n+1}=(n+1)(1+a_n)$ We have $$P_n=\prod_{i=1}^n(1+a_i^{-1})$$ then

$$ P_n =\prod_{i=1}^n (1 + \frac{1}{a_i}) =\left (1 + \frac{1}{n(1+a_{n-1})}\right )\prod_{i=1}^{n-1} (1+a_i^{-1} ) =\left (\frac{n(1+a_{n-1}) + 1}{n(1+a_{n-1})}\right )\prod_{i=1}^{n-1} (1+a_i^{-1} ) =\left (\frac{n(1+a_{n-1}) + 1}{n(1+a_{n-1})}\right ) \left (\frac{(n-1)(1+a_{n-2}) + 1}{(n-1)(1+a_{n-2})}\right )\prod_{i=1}^{n-2} (1+a_i^{-1} ) = \ldots \approx \frac{n! + \prod_{i=1}^n a_i}{n! + \prod_{i=1}^n a_i} = 1 $$

answered Jun 20 '14 at 18:26

Morteza Shahriari Nia

191

looks like karolis's answer is correct – Morteza Shahriari Nia Jun 20 '14 at 18:29
I am sure that the answer is e – RE60K Jun 20 '14 at 18:31
Your '...' in the last line is completely wrong - you're saying, in essence, that $(a_n+b_n)(a_{n-1}+b_{n-1})\cdots(a_1+b_1) = a_na_{n-1}\cdots a_1 + b_nb_{n-1}\cdots b_1$. (Where you pass from the complicated product to the simple one) – Steven Stadnicki Jun 20 '14 at 19:04
agreed, too much simplification – Morteza Shahriari Nia Jun 20 '14 at 19:08

Limit of a sequence $a_1=1;a_{n+1}=(n+1)(1+a_n)$

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