Problem on limit of a sequence

Question

I got stuck on a problem recently which apparently leads to bizarre conclusions..

Let $a_1=1$, and $a_n=n(a_{n-1}+1)$ for all $n={2,3,...}$ Define $$P(n)=\Biggl(1+\frac{1}{a_1}\Biggl)\Biggl(1+\frac{1}{a_2}\Biggl)...\Biggl(1+\frac{1}{a_n}\Biggl)$$

Compute: $$\lim_{n\to\infty}P(n)$$ The way I did it was..i simplified $P(n)$ to get $P(n)=\frac{a_{n+1}}{(n+1)!}$

Now, rearranging the expression I got $$\frac{1}{n}=\frac{a_{n-1}+1}{a_n}$$ Let $\lim_{n\to\infty}a_n=\Delta$, then taking the limit as n$\to\infty$ of the above expression gives $0=\frac{\Delta+1}{\Delta}$ leading to $\Delta=-1$ but clearly, $\{a_n\}$ is an increasing sequence..where am I wrong?

This line of reasoning also leads to the required limit being 0..which again is obviously wrong..

Edit:I want to know what's wrong in what I've done..and I haven't seen this approach elsewhere. I know there are solutions to this already on this website..but the approach there is different..

Does this answer your question? Let $a_1=1$ and $a_n=n(a_{n-1}+1)$ for $n=2,3,...$. — LHF, Mar 03 '20 at 07:32
No..I've seen that before..I wanted to know what's wrong in my understanding.. — thornsword, Mar 03 '20 at 07:35

score 0 · Accepted Answer · answered Mar 03 '20 at 07:29

0

$a_n >n$ for all $n$ so $a_n$ does not converge to a finite limit $\Delta$. Your mistake is in assuming that $\Delta$ is finite.

answered Mar 03 '20 at 07:29

Kavi Rama Murthy

311,013

okay..but don't we assume the limit to be some number say alpha and continue in most problems? – thornsword Mar 03 '20 at 07:39
1

@binarybitarray, you first have to prove it is convergent first. – LHF Mar 03 '20 at 07:39
Okay, if possible can you give links to an example problem so that I understand it better? – thornsword Mar 03 '20 at 07:41
Example: https://math.stackexchange.com/questions/3552900/show-that-lim-n-to-infty-x-n-exists-and-find-it – LHF Mar 03 '20 at 07:42

Problem on limit of a sequence

1 Answers1

Linked