# Advanced Calculus Proof-limit of a sequence.

1. Sep 28, 2008

### emira

1. The problem statement, all variables and given/known data
Suppose that the sequence {a_n} converges to A. Define the sequence {b_n} by:

b_n = (a_n +a_(n+1) )/2

Does the sequence {b_n} converge? If so, specify the limit and prove your conclusion. Otherwise, give an example when this is not true.

2. Relevant equations

3. The attempt at a solution

Me and some class mates tried to use the sandwich theorem, but are not sure if it applies here:

Suppose that {a_n}, {k_n}, and {c_n} are three sequences. NOTE: k_n = a_(n+1). Suppose there exists an element n from N, such that

a_n≤k_n≤c_n

Suppose that a_n and c_n converge to A, then, by the limit definition, there exists an n_2 such that |a_n-A|<ε

the same thing for c_n, there exists an n_3 such that |c_n-A|<ε

n* max {n_1, n_2, n_3}

A-ε≤a_(n )≤A+ε
A-ε≤c_(n )≤A+ε
then A-ε≤a_(n )≤k_n≤c_n≤A+ε so that means that K_n converges to A, but k_n is actually
a_(n+1), so then a_(n+1) converges to A.

Now, since b_n =( a_n +a_(n+1))/2

take the limit of b_n and by limit preperties you find that the limit of b_n is A.

I dont know if this is the right solution for this problem, we also thought about representing the a_n as a sequence that converges to A, such as a_n =A, but Im not sure since it seems we are asked to prove this generally.

Any hints would be highly appreciated,

thank you very much,

Emira!

2. Sep 28, 2008

### tiny-tim

Hi emira!

Just start "given ε, ∃ n such that …"

3. Sep 28, 2008

### HallsofIvy

Staff Emeritus
Part of that, at least, is easy: If {bn} converges, then its limit is easy to get:
$$\lim{n\rightarrow\infty}\frac{a_n+ a_{n+1}}{2}= \frac{\lim_{n\rightarrow\infty} a_n}{2}+ \frac{\lim_{n\rightarrow\infty} a_{n+1}}{2}$$
and those last two limits are given.

As for proving that it does converge, since we know {an} converges to A, given any $\epsilon> 0$, there exist N such that if n> N, $|a_n- A|< \epsilon$. Of course, if n> N, so is n+1 so the same is true of an+1!

4. Sep 28, 2008

Since

$$\lim_{n \to \infty} a_n = A$$

then it is also true that

$$\lim_{n \to \infty} a_{n=1} = A$$

The proof that $$\{b_n\}$$ converges is simply an exercise in the algebra of limits - no need for an $$\epsilon, \delta$$ proof (unless the original question required such).