Find Limit of na(n): Help Requested

[Tolya]

Please, help to find limit:
\lim_{n \rightarrow \infty} na(n), where
a(1)=1;<br /> a(n+1)=\frac{a(n)}{1+\left|sin(a(n))\right|}
Thanks for any ideas!

 

[Avodyne]

If the limit exists, then a(n) is going to zero at large n, which can be used to simplify the right-hand side.

Also, at large n, a finite-difference equation can be well approximated by a differential equation.

 

[Tolya]

Thanks. With the help of your post, Avodyne, I found that limit equals 1. Is it correct?
Sorry, but I'm not sure.

 

[Avodyne]

Yes, this is correct!

 

[Tolya]

But actually, I didn't use differential equation :)
\frac{1}{a_{n+1}}=\frac{1+\left|sin(a_n)\right|}{a_n}
It's easy to show that a_n is going to zero at large n, but remains postive. So:
\frac{1}{a_{n+1}}=\frac{1+a_n-\frac{1}{6}a_n^3+o(a_n^3)}{a_n} =\frac{1}{a_n}+1+b_n, where b_n \rightarrow 0, when n \rightarrow \infty.
Use this we can obtain:
\frac{1}{a_{n+1}}=\frac{1}{a_1}+n+b_1+b_2+...+b_n
\frac{1}{na_{n+1}}=\frac{1}{na_1}+1+\frac{b_1+b_2+...+b_n}{n}
When n is going to infinity, we have: (using well-known \lim_{n \rightarrow \infty}\frac{b_1+b_2+...+b_n}{n}=0, where each of b is going to zero with large n)
\lim_{n \rightarrow \infty} \frac{1}{na_{n+1}}=0+1+0=1

In any case, thank you very much, Avodyne. :)

 

[Avodyne]

Very nice. Your analysis is more rigorous than mine was.