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August 21st, 2011, 10:30 AM   #1
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Limit Superior and Limit Inferior

Prove that for any positive sequence a_n of real numbers
lim inf (a _(n+1) / a_n) <= lim inf (a_n)^(1/n) <= lim sup (a_n)^(1/n)
<= lim sup(a_(n+1) / a_n).
Give examples where equality does not hold.

Is lim sup always >= lim inf? I am having trouble understanding these concept and proving things about them without specific numbers.
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August 21st, 2011, 01:10 PM   #2
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Re: Limit Superior and Limit Inferior

The difficulty you are having with this concept may come from the definition you were given for limsup and liminf. There are actually several alternative (equivalent) ways to define this notion. Usually a textbook gives one as a definiton, and one or two others are given as characterisations.
See p. 38 of Real Analysis by H. L. Royden. Another method may be using subsequences.
Further, using the inf-sup definition a handy inequality is given in p 55 of the Elements of Real Analysis by Bartle, which gives you enough clue to solve this.
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August 21st, 2011, 01:43 PM   #3
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Re: Limit Superior and Limit Inferior

Thanks for the answer, but I don't have access to those books. Could you tell me the definitions or inequality you are talking about?
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August 22nd, 2011, 10:59 AM   #4
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Re: Limit Superior and Limit Inferior

Quote:
Originally Posted by veronicak5678
Thanks for the answer, but I don't have access to those books. Could you tell me the definitions or inequality you are talking about?
Ok! But it's going to give us both a very hard time due to my LaTeX illiteracy!
H. L. Royden has given the following definition for lim sup and lim inf:
for a sequence <x_n> of real numbers
lim sup xn:= inf_n (sup x_k) for k no lesser than n.
Lim inf is defined in the same fashion.
Now, you may prove the following:
1.
-a real no. L is the limsup of <xn> iff
i. For e>0 there is some n with xm<L+e for all m >= n, and
2. Given both e>0 and n, there is some m, m>=n and xm>l-e.
- +oo is the lim sup of our sequence iff it is a cluster point of it
- -oo is the lim sup of <xn> iff it is its limit.
Now, for lim inf we have the following useful propositions:
limsup (-xn)= -liminf xn
liminf xn<= limsup xn, equality holds iff xn converges, then they both equal that limit.
More importantly,
lim sup and lim inf are the largest, and smallest cluster points of a sequence respectively.
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August 22nd, 2011, 11:07 AM   #5
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Re: Limit Superior and Limit Inferior

Some useful inequalities are:
limsup xn + lim inf yn <= limsup (xn+yn)<= limsup xn+ limsup yn
limsup (xn.yn)<= limsupxn.limsup yn
also check the definition of exponentiation for rationals using supremum.
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