Cfrgtkky

My problem is, that I have sequence and I am supposed to make a limit.

My sequence is $ a_{n+1}=frac{1}{2}cdot (a_{n}+frac{1}{a_{n}}), a_{1}=2$.

I know list first few terms:

$ left { 2, frac{5}{4},frac{41}{40},frac{3281}{3280} right } $

but I am not able to solve it. Can anyone please help me? Thanks.

I can't say that this is the most efficient way of tackling these kinds of problems, but here is a rough plan of attack for recurrences:

1) Compute some initial terms of the sequence (if feasible) in order to have a grip on some basic initial behavior.

2) Based on what you know about the recurrence, or what you observe in your computations in step 1, see if there is a general property you can prove or are willing to conject about the sequence. In the best case, you may even be able to guess the limit.

3) Leverage theorems at your disposal to actually prove your conjectures correct from step 2.

In this particular problem, you have computed some initial terms, and we can now try and make some general statements. Clearly, $a_n$ will always be positive. Less clearly, but still very easy to prove, is that $a_nleq 2$. Even less clearly, and slightly harder to prove, is that your sequence $a_n$ is decreasing.

Ah ha! A decreasing sequence bounded below is very special. We have a theorem saying that the limit of such a sequence must exist. Since this limit exists, give it a name, say $lim a_n = a$.

This final fact is going solve our entire problem. Take your recurrence $a_{n+1}=1/2(a_n+1/a_n)$ and take the limit of both sides. This will be a single algebraic equation in $a$, which you can (in principle) solve, and use the earlier properties of the sequence to narrow down which solution of the equation (if there are multiple ones) the real answer is.

First assume that the sequence converges. Then we let $a := lim_{n to infty} a_n$ be the limit. Further assume that $a neq 0$. Now we obtain that
$$ a = lim_{n to infty} a_{n + 1} = lim_{n to infty} frac 1 2 (a_n + frac{1}{a_n}) = frac 1 2 (a + frac{1}{a}).$$
This expression is equivalent to $a^2 = 1$ which implies that $a in {1, -1}$ under the assumption that the limit exists. Now we are left to check whether the sequenceconverges and which of the two possible limits is the right one. Recall that every sequence that is decreasing and bounded from below converges. So try to show that:

1. 
   $(a_n)_{n in mathbb N}$ is decreasing, i.e., $a_{n + 1} leq a_n$ for all $n in mathbb N$.


2. 
   $(a_n)_{n in mathbb N}$ is bounded from below. In particular, you should be able to verify that $a_n geq 1$ for all $n in mathbb N$.

If you show these two things, the calculation above is justfied and $a_n geq 1$ for all $n in mathbb N$ yields that $a geq 1$ and thus $a = 1$ by the calculation above. I hope it helps you :)