# Cauchy sequence question

1. ### Unassuming

169
Let a_1 and a_2 be arbitrary real number that are not equal. For $$n \geq 3$$, define a_n inductively by,

$$a_n = \frac{1}{2} (a_{n-1}+a_{n+2} )$$

I cannot get the result that the book gets. I proceed,

$$a_{n+1} - a_{n} = \frac{1}{2}(a_n + a_{n-1} ) - \frac{1}{2} (a_{n-1} + a_{n-2} ) = \frac{1}{2} ( a_n - a_{n-2} )= \frac{1}{2}(a_n + a_{n-1} )$$

$$a_{n+1} - a_n = \frac{-1}{2} (a_n - a_{n-1} )$$

Any help for me?

2. ### Dick

25,851
You've got some typos in there that are making this really confusing. The original recursion is a_n=(1/2)*(a_n-1+a_n-2), right? That says 2*a_n-a_n-1=a_n-2. I agree with the book.

3. ### Unassuming

169
I can't see it. I've done some algebra and I still can't get anywhere from your last step.

4. ### Dick

25,851
You've got a_n+1-a_n=(1/2)(a_n-a_n-2), right? Put the expression for a_n-2 into that. If you still aren't getting it show your work.

5. ### Unassuming

169
I got it. Thanks Dick

6. ### HallsofIvy

40,514
Staff Emeritus
Here's a "cheat":
Since $\sum x^n/n!$ is the Taylor's series for ex, 1+ 1/2!+ 1/3!+ ... converges to e. Since it converges, it is a Cauchy sequence!

Or you could just note that the numeric series converges by the ratio test. Again, since it converges, it is a Cauchy sequence.

But I doubt that either of those answers is what was wanted!

7. ### Unassuming

169
Those are nice things to put the problem into perspective though. Sometimes I (and hopefully others) get lost in the algebra of the problem.