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November 17th, 2016, 01:15 PM   #1
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proof of limit

I know that I have two series that have limits, an and bn
and I have two groups {an>=bn) and {bn>=an), n is natural. I know about the groups that they aren't bounded.
I have to prove that limit an=limit bn. Thanks.

Last edited by skipjack; November 17th, 2016 at 03:09 PM.
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November 17th, 2016, 01:34 PM   #2
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duplicate post ...

http://mymathforum.com/complex-analy...prove-lim.html
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November 17th, 2016, 01:54 PM   #3
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Duplicate deleted.
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November 17th, 2016, 07:14 PM   #4
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Quote:
Originally Posted by moshelevi33 View Post
I know that I have two series that have limits, an and bn
and I have two groups {an>=bn) and {bn>=an), n is natural. I know about the groups that they aren't bounded.
I have to prove that limit an=limit bn. Thanks.
I think what you mean is that, for some infinite number of values of "n", an<= bn and, for another infinite set of values of "n", bn<=an.

Use "proof by contradiction". Suppose the two limits are not equal. Then, calling the two limits A and B either A> B or B> A.

case I: A> B. Let d= A- B. Then by definition of "limit" there exist N such that |an- A|< d/10. But there exist an infinite number of n such that bn> an so there exist an infinite number of n for which bn> an> A- (A- B)/2= (A+ B)/2. But that contradicts the fact that bn converges to B.

case 2: B> A. Just reverse "A" and "B" in case (1).
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November 18th, 2016, 06:39 AM   #5
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In general, a numerical series, $\displaystyle \sum a_n$, with all $\displaystyle a_m> 0$, converges as long as $\displaystyle \frac{a_{n+1}}{a_n}$ converges to a number less than 1 (the "ratio test"). In particular, the power series, $\displaystyle \sum a_mx^n$ converges (absolutely) if and only if $\displaystyle \left|\frac{a_{n+1}x^{n+1}}{a_nx^n}\right|= \left|\frac{a_{n+1}}{a_n}\right||x|$ converges to a number less than 1. Further, if $\displaystyle \left|\frac{a_{n+1}}{a_n}\right|$ converges to a non-zero number, A, then we must have $\displaystyle A|x|< 1$ so that $\displaystyle |x|> \frac{1}{A}$.

If $\displaystyle \left|\frac{a_{n+1}}{a_n}\right|$ converges to A> 0 then the radius of convergence is 1/A. if it converges to 0, the series has infinite radius of convergence (it converges for all x) and if it does not converge (goes to 0) the radius of convergence is 0 (it converges only for x = 0).
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Last edited by skipjack; November 18th, 2016 at 02:10 PM.
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November 18th, 2016, 11:47 AM   #6
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Let an and bn be convergent infinite sequences: a1,a2,..... & b1,b2,...
an $\displaystyle \geq$ bn $\displaystyle \rightarrow$ A $\displaystyle \geq$ B
bn $\displaystyle \geq$ an $\displaystyle \rightarrow$ B $\displaystyle \geq$ A
A=B
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November 18th, 2016, 12:41 PM   #7
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I think that's a fail. It might be possible to rescue it though.

Last edited by v8archie; November 18th, 2016 at 12:43 PM.
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November 19th, 2016, 06:27 AM   #8
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Let an and bn be convergent infinite sequences, an=a1,a2,..... and bn=b1,b2,.....,

Let a'n and b'n be infinite subsequences of an and bn respectively st an' $\displaystyle \geq$ bn'.
Let a''n and b''n be infinite subsequences of an and bn respectively st bn''$\displaystyle \geq$ an''.
an' $\displaystyle \geq$ bn' $\displaystyle \rightarrow$ A' $\displaystyle \geq$ B'
bn''$\displaystyle \geq$ an'' $\displaystyle \rightarrow$ B'' $\displaystyle \geq$ A"

But A'=A''=A and B'=B''=B $\displaystyle \rightarrow$ A=B.

What has been proven is that it is only possible to make the given selection of subsets if A=B.


Notes:
an used to denote sequence and individual member of sequence for convenience.
If an infinite set has one cluster point, any infinite sub-set has the same cluster point
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November 19th, 2016, 07:30 AM   #9
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That would work if you actually referred to the theorem at the heart of the proof. Without that, it's just a bunch of unproven claims.

When you try to prove something, you get nothing for brevity. All the credit is for clarity.

Studious avoidance of normal notation, $\LaTeX$ and any attempt to clearly define your terms is one reason that everything you write appears wrong even if there is a core of truth.

Edit: perhaps your footnote is the thing that I'm looking for, but it uses undefined terms and, as written, is incorrect anyway.

Last edited by v8archie; November 19th, 2016 at 07:32 AM.
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November 21st, 2016, 11:25 AM   #10
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Quote:
Originally Posted by zylo View Post
Let an and bn be convergent infinite sequences, an=a1,a2,..... and bn=b1,b2,.....,

Let a'n and b'n be infinite subsequences of an and bn respectively st an' $\displaystyle \geq$ bn'.
Let a''n and b''n be infinite subsequences of an and bn respectively st bn''$\displaystyle \geq$ an''.
an' $\displaystyle \geq$ bn' $\displaystyle \rightarrow$ A' $\displaystyle \geq$ B'
bn''$\displaystyle \geq$ an'' $\displaystyle \rightarrow$ B'' $\displaystyle \geq$ A"

But A'=A''=A and B'=B''=B $\displaystyle \rightarrow$ A=B.

What has been proven is that it is only possible to make the given selection of subsets if A=B.


Notes:
an used to denote sequence and individual member of sequence for convenience.
If an infinite set has one cluster point, any infinite sub-set has the same cluster point
Example:

a: 1,2,3,4,5....an..., and an st it converges to A
b: 2,4,6,8,10....bn..., and bn st it converges to B

a'=1,3,5,7,.....a'n...., infinite subset of a
b'=2,4,6,8,....b'n...., infinite subset of b
b'n $\displaystyle \geq$ a'n $\displaystyle \rightarrow$ B' $\displaystyle \geq$ A'

a''=7,9,11,....a''n..., infinite subset of a
b"=2,6,10.....b''n..., infinite subset of b
a''n $\displaystyle \geq$ b''n $\displaystyle \rightarrow$ A" $\displaystyle \geq$ B''

Lim a'n=A', Lim b'n=B', Lim a''n=A'', Lim b''n= B''
B' $\displaystyle \geq$ A'
A" $\displaystyle \geq$ B''
A'=A''=A
B'=B''=B
B $\displaystyle \geq$ A
A $\displaystyle \geq$ B
A=B

The point is you can't pick subsequences (subsets) of a and b such that:
b'n $\displaystyle \geq$ a'n and a''n $\displaystyle \geq$ b''n, n infinite, unless A = B. You can start off that way, as I did, but eventually it's only possible for infinite n if A=B.
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