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_{n }for all n, show A< or = B.

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- Mar 16th 2013, 04:58 PMamywReal analysis; sequences and limits
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_{n }for all n, show A< or = B. - Mar 16th 2013, 05:36 PMProve ItRe: Real analysis; sequences and limits
Nah, no need for contradiction. If $\displaystyle \displaystyle \begin{align*} a_n < b_n \end{align*}$, then $\displaystyle \displaystyle \begin{align*} b_n = a_n + c_n \end{align*}$, where $\displaystyle \displaystyle \begin{align*} c_n > 0 \end{align*}$ for all n, which must go to some positive value (C) since B is finite. Then

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

But we also know this limit is B, therefore B = A + C, and thus $\displaystyle \displaystyle \begin{align*} B \geq A \end{align*}$. - Mar 16th 2013, 07:36 PMamywRe: Real analysis; sequences and limits
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, - Mar 16th 2013, 07:43 PMProve ItRe: Real analysis; sequences and limits
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. - Mar 16th 2013, 09:06 PMjohngRe: Real analysis; sequences and limits
Hi,

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

Attachment 27567 - Mar 17th 2013, 04:04 AMamywRe: Real analysis; sequences and limits
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. - Mar 17th 2013, 05:18 AMPlatoRe: Real analysis; sequences and limits
This is a well-known problem used to teach a particular way of proof.

Suppose that $\displaystyle b<a$ then $\displaystyle b<\frac{a+b}{2}<a$

So let $\displaystyle \varepsilon = \frac{{a - b}}{2} > 0$. Now note that $\displaystyle a - \varepsilon = \frac{{b + a}}{2}\;\& \;b + \varepsilon = \frac{{b + a}}{2}.$

That means $\displaystyle \left( {b - \varepsilon ,b + \varepsilon } \right) \cap \left( {a - \varepsilon ,a + \varepsilon } \right) = \emptyset $.

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

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

Surely you can see a contradiction there? - Mar 17th 2013, 07:18 AMjohngRe: Real analysis; sequences and limits
Hi Amy,

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

Attachment 27576 - Mar 17th 2013, 05:39 PMamywRe: Real analysis; sequences and limits
Thanks, this makes perfect sense to me now that I've had some sleep.