November 17th, 2016, 01:15 PM  #1 
Newbie Joined: Nov 2016 From: Uk Posts: 11 Thanks: 0  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. 
November 17th, 2016, 01:34 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,321 Thanks: 1142  
November 17th, 2016, 01:54 PM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,411 Thanks: 862 Math Focus: Elementary mathematics and beyond 
Duplicate deleted.

November 17th, 2016, 07:14 PM  #4  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,282 Thanks: 570  Quote:
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).  
November 18th, 2016, 06:39 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,282 Thanks: 570 
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 $\displaystyle \left\frac{a_{n+1}x^{n+1}}{a_nx^n}\right= \left\frac{a_{n+1}}{a_n}\rightx$ converges to a number less than 1. Further, if $\displaystyle \left\frac{a_{n+1}}{a_n}\right$ converges to a nonzero number, A, then we must have $\displaystyle Ax< 1$ 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). Last edited by skipjack; November 18th, 2016 at 02:10 PM. 
November 18th, 2016, 11:47 AM  #6 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 918 Thanks: 72 
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 
November 18th, 2016, 12:41 PM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,454 Thanks: 2126 Math Focus: Mainly analysis and algebra 
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. 
November 19th, 2016, 06:27 AM  #8 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 918 Thanks: 72 
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 subset has the same cluster point 
November 19th, 2016, 07:30 AM  #9 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,454 Thanks: 2126 Math Focus: Mainly analysis and algebra 
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. 
November 21st, 2016, 11:25 AM  #10  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 918 Thanks: 72  Quote:
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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