Real analysis; sequences and limits

**amyw** · March 16th 2013, 04:58 PM

I think I need to do this by contradiction, but I'm stuck.

Prove:

If the sequence <a_n> has the limit A, the sequence <b_n> has the limit B, and a_n<b_nfor all n, show A< or = B.

**Prove It** · March 16th 2013, 05:36 PM

Nah, no need for contradiction. If $\displaystyle \begin{align*} a_n < b_n \end{align*}$ , then $\displaystyle \begin{align*} b_n = a_n + c_n \end{align*}$ , where $\displaystyle \begin{align*} c_n > 0 \end{align*}$ for all n, which must go to some positive value (C) since B is finite. Then

$\displaystyle \begin{align*} \lim_{n \to \infty} b_n &= \lim_{n \to \infty} \left( a_n + c_n \right) \\ &= \lim_{n \to \infty} a_n + \lim_{n \to \infty} c_n \\ &= A + C \end{align*}$

But we also know this limit is B, therefore B = A + C, and thus $\displaystyle \begin{align*} B \geq A \end{align*}$ .

**amyw** · March 16th 2013, 07:36 PM

How do you know B is finite? Also, how do you know the c sub n has limit C?

In class, we have done these type of proof using the definition of converge. I can follow this, mostly, but I don't think I'm allowed to do the proof this way.

Thanks,

**Prove It** · March 16th 2013, 07:43 PM

If a sequence has a limiting value of B, then B is a number and is clearly finite.

In order for all the b terms to be greater than the a terms, then it's the same as adding some value to the a terms. Since the a terms go to some value A, in order for the b terms to also go to some value, then the remaining c terms also have to go to some value.

**johng** · March 16th 2013, 09:06 PM

Hi,

In direct answer to your question about proof by contradiction, I offer the following:

$Real analysis; sequences and limits-mhfsequencelimita.png$

**amyw** · March 17th 2013, 04:04 AM

This is helpful, but did you let epsilon = -L/2? which is positive because L is less than zero. But where did you get -L/2? I'm guessing it's from some scratch work that should be obvious to me, but isn't.

Also, we haven't proved or used that the limit of the difference two convergent sequences is the the difference of the limits. I know it's true, but since we haven't proved it I can't use it.

**Plato** · March 17th 2013, 05:18 AM

Originally Posted by amyw

This is helpful, but did you let epsilon = -L/2? which is positive because L is less than zero. But where did you get -L/2? I'm guessing it's from some scratch work that should be obvious to me, but isn't.
Also, we haven't proved or used that the limit of the difference two convergent sequences is the the difference of the limits. I know it's true, but since we haven't proved it I can't use it.

This is a well-known problem used to teach a particular way of proof.
Suppose that $b<a$ then $b<\frac{a+b}{2}<a$

So let $\varepsilon = \frac{{a - b}}{2} > 0$ . Now note that $a - \varepsilon = \frac{{b + a}}{2}\;\& \;b + \varepsilon = \frac{{b + a}}{2}.$
That means $\left( {b - \varepsilon ,b + \varepsilon } \right) \cap \left( {a - \varepsilon ,a + \varepsilon } \right) = \emptyset$ .

But because $(b_n)\to b$ almost all of the terms are in $\left( {b - \varepsilon ,b + \varepsilon } \right)$ .

AND because $(a_n)\to a$ almost all of the terms are in $\left( {a - \varepsilon ,a + \varepsilon } \right)$ .

Surely you can see a contradiction there?

**johng** · March 17th 2013, 07:18 AM

Hi Amy,

I hope the following is yet more "insight" into your problem:

$Real analysis; sequences and limits-mhfsequencelimit2.png$

**amyw** · March 17th 2013, 05:39 PM

Thanks, this makes perfect sense to me now that I've had some sleep.

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