{a_n}\to A and define {b_n}

**dwsmith** · June 18th 2011, 07:18 PM

$\{a_n\}_{n=1}^{\infty}\to A$ and define $b_n=\frac{a_n+a_{n+1}}{2}, \ \ \forall n$

Prove $\{b_n\}_{n=1}^{\infty}\to A$

Is it as simple as $b_n=\frac{A+A}{2}=A$

Therefore, $\{b_n\}_{n=1}^{\infty}\to A$

**TheEmptySet** · June 18th 2011, 08:58 PM

Originally Posted by dwsmith

$\{a_n\}_{n=1}^{\infty}\to A$ and define $b_n=\frac{a_n+a_{n+1}}{2}, \ \ \forall n$

Prove $\{b_n\}_{n=1}^{\infty}\to A$

Is it as simple as $b_n=\frac{A+A}{2}=A$

Therefore, $\{b_n\}_{n=1}^{\infty}\to A$

You need a little more rigor.

Since $a_n \to A$ Pick $N \in \mathbb{N}$ such that for all $n > N$

$|a_n-A|< \frac{\epsilon}{2}$

Now what is

$|b_n-A|=\bigg| \frac{a_n+a_{n+1}}{2}-\frac{2A}{2}\bigg|=...$

**dwsmith** · June 19th 2011, 07:21 AM

I don't know why you picked $\frac{\epsilon}{2}$ and I don't know where to go from the $=\cdots$ .

Can you elaborate further.

**Plato** · June 19th 2011, 07:48 AM

Originally Posted by dwsmith

I don't know why you picked $\frac{\epsilon}{2}$ and I don't know where to go from the $=\cdots$ .

Actually we could pick just ${\epsilon}$ if you note that
$\left| {\frac{{a_n + a_{n + 1} }}{2} - A} \right| \leqslant \frac{1}{2}\left| {a_n - A} \right| + \frac{1}{2}\left| {a_{n + 1} - A} \right|~.$

**Deveno** · June 19th 2011, 08:46 AM

Originally Posted by dwsmith

I don't know why you picked $\frac{\epsilon}{2}$ and I don't know where to go from the $=\cdots$ .

Can you elaborate further.

you can rewrite this difference as the sum of two differences, and use a version of the triagle inequality.....

the "idea" behind this is that any convergent series is Cauchy, so for large enough n, two successive terms are as close as we like (this means something in terms of epsilon), so the "average" of the two successive terms (for large enough n) is going to be close to the limit as well.

**dwsmith** · June 19th 2011, 06:13 PM

Originally Posted by Plato

Actually we could pick just ${\epsilon}$ if you note that
$\left| {\frac{{a_n + a_{n + 1} }}{2} - A} \right| \leqslant \frac{1}{2}\left| {a_n - A} \right| + \frac{1}{2}\left| {a_{n + 1} - A} \right|~.$

Ok, so from that, we can say:

$\frac{1}{2}|a_n-A|<\epsilon\Rightarrow |a_n-A|<2\epsilon\Rightarrow |a_n-A|<\epsilon_2$

Therefore, $\{a_n\}\to A$ and the same argument for $a_{n+1}$ .

From this, we can then use the definition of b_n with both sequences being A, correct?

**Deveno** · June 19th 2011, 07:12 PM

since {an}-->A, for any ε > 0, there IS some N with |an - A| < ε for all n > N. so 1/2 of that is less than ε/2.

if n > N, surely n+1 > N as well, thus |a(n+1) - A| < ε, too. there is no need for a "secondary" epsilon:

$\left|\frac{a_n + a_{n+1}}{2} - A}\right|\leq\frac{1}{2}|a_n-A|+\frac{1}{2}|a_{n+1}-A| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

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