# {a_n}\to A and define {b_n}

• Jun 18th 2011, 07:18 PM
dwsmith
{a_n}\to A and define {b_n}
$\{a_n\}_{n=1}^{\infty}\to A$ 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, $\{b_n\}_{n=1}^{\infty}\to A$
• Jun 18th 2011, 08:58 PM
TheEmptySet
Re: {a_n}\to A and define {b_n}
Quote:

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

$|b_n-A|=\bigg| \frac{a_n+a_{n+1}}{2}-\frac{2A}{2}\bigg|=...$
• Jun 19th 2011, 07:21 AM
dwsmith
Re: {a_n}\to A and define {b_n}
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.
• Jun 19th 2011, 07:48 AM
Plato
Re: {a_n}\to A and define {b_n}
Quote:

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
$\left| {\frac{{a_n + a_{n + 1} }}{2} - A} \right| \leqslant \frac{1}{2}\left| {a_n - A} \right| + \frac{1}{2}\left| {a_{n + 1} - A} \right|~.$
• Jun 19th 2011, 08:46 AM
Deveno
Re: {a_n}\to A and define {b_n}
Quote:

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.
• Jun 19th 2011, 06:13 PM
dwsmith
Re: {a_n}\to A and define {b_n}
Quote:

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?
• Jun 19th 2011, 07:12 PM
Deveno
Re: {a_n}\to A and define {b_n}
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:

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