limit of a sequence -> possibly squeeze theorem

http://img96.imageshack.us/img96/8606/qqqqbj.jpg

so what i'm thinking is that you let

x_n+3 + x_n+2<=x_n+2 + x_n+1 <= x_2 + x_1

then show that the first term can eventually go to the last term

squeezing the term into the middle

but the problem with this is i dont see where the 1/3 comes from

any ideas?

Re: limit of a sequence -> possibly squeeze theorem

Quote:

Originally Posted by

**Angela11** so what i'm thinking is that you let

x_n+3 + x_n+2<=x_n+2 + x_n+1 <= x_2 + x_1

What do you mean by "let"? One uses the word "let" during a definition: e.g., let w = 6. Why do you think these inequalities are true?

Quote:

Originally Posted by

**Angela11** any ideas?

One way is to find an explicit formula for . It is convenient to define and so that xₙ = (aₙx₁ + bₙx₂) / 2ⁿ⁻². Then it is possible to write recurrence relations and initial conditions on aₙ and bₙ, solve them and find the limits of aₙ / 2ⁿ⁻² and bₙ / 2ⁿ⁻².

Re: limit of a sequence -> possibly squeeze theorem

The way to do simple linear recursions is to try . This is very much analogous to ODEs, where you'd try , seeing what f must be, and then using superposition to add together the different possibilities for f.

Solve: .

Solution: Try for some constants

From get , so .

Solving that gives or . Now just like with ODEs, put them together to get the general form of the solution:

.

Note the required 2 unknown constants for a degree 2 linear recursion.

When you plug that in to the recursion equation, it works. It's just algebra - I won't work it out here. Thus we have the generic solution to that recursion.

Just like with ODEs, we use the initial conditions to get the specific solution:

, and . Thus

, and , which add to give , so .

To solve the problem, we don't even need to bother to find , though it would be simple to do.