# Help to find limit

1. Nov 8, 2007

### Tolya

$$\lim_{n \rightarrow \infty} na(n)$$, where
$$a(1)=1; a(n+1)=\frac{a(n)}{1+\left|sin(a(n))\right|}$$
Thanks for any ideas!

2. Nov 8, 2007

### 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.

3. Nov 8, 2007

### Tolya

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

Last edited: Nov 8, 2007
4. Nov 8, 2007

### Avodyne

Yes, this is correct!

5. Nov 8, 2007

### 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. :)

6. Nov 8, 2007

### Avodyne

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