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Science Forum Index » Mathematics Forum » Convergence of the geometric series
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Message |
| Dave Parker |
 Posted: Sat Feb 12, 2005 3:26 pm |
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Guest
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It is well-known that the geometric series 1+a+a^2+a^3... for |a|<1
sums
to 1/(1-a).
The proof of this follows from the following fact applied to the
partial sum:
lim a^n = 0 if |a|<1.
n->oo
Everybody considers this last statement "obvious", but I have never
seen an explicit proof of it.
By applying the definition of limit, the only proof I could find
requires the use of logarithms. (Given an epsilon, you have to find an
n s.t. a^n < epsilon, etc. Logs enter the picture naturally.)
Here is the problem. In most analysis books the development of
logarithms, exponentials, etc. depend on various facts about series,
which when you go through the layers of convergence theorems eventually
end up depending on the convergence of the geometric series. So there
is a chicken and egg problem.
What is an elementary proof of
lim a^n = 0 if |a|<1.
n->oo
that doesn't depend on logarithms or other higher functions?
Thanks.
Dave |
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| Stuart M Newberger |
| Posted: Sat Feb 12, 2005 4:06 pm |
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"Dave Parker" <dave3852@yahoo.com> wrote in message
news:1108240007.618379.45440@g14g2000cwa.googlegroups.com...
Quote: It is well-known that the geometric series 1+a+a^2+a^3... for |a|<1
sums
to 1/(1-a).
The proof of this follows from the following fact applied to the
partial sum:
lim a^n = 0 if |a|<1.
n->oo
Everybody considers this last statement "obvious", but I have never
seen an explicit proof of it.
By applying the definition of limit, the only proof I could find
requires the use of logarithms. (Given an epsilon, you have to find an
n s.t. a^n < epsilon, etc. Logs enter the picture naturally.)
Here is the problem. In most analysis books the development of
logarithms, exponentials, etc. depend on various facts about series,
which when you go through the layers of convergence theorems eventually
end up depending on the convergence of the geometric series. So there
is a chicken and egg problem.
What is an elementary proof of
lim a^n = 0 if |a|<1.
n->oo
that doesn't depend on logarithms or other higher functions?
Thanks.
Dave
Well the sequence is decreasing and positive and so must converge (monotone
convergernce property of real numbers).If its limit is c then since
a_(n+1)=aa_n and both a_n and a_(n+1) converge to c we have c=ac or c(1-a)=0
..Since a is not 1 we conclude by division that c=0.You need a good advanced
calculus book.A previous generation or too learned from Rudin ,W "
Principles of Mathematical Analysis".Its still available but way
overpriced -try the marketplace.
Regards,Stuart M Newberger |
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| JEMebius |
| Posted: Sat Feb 12, 2005 4:26 pm |
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One proves rather easily that every increasing exponential function
grows faster than any increasing linear function.
The proof consists of algebraic reasonings that the theorem holds for
integer and rational arguments, and of a reasoning assuming continuity
that it holds for the complete real line. For your purpose you only need
integer arguments, so one can drop the second and third parts of the proof.
Therefore, every decreasing exponential function decreases faster than
the function x => 1/x. And 1/x can be made arbitrary small by choosing x
large enough. There you are!
Cheers: Johan E. Mebius
Dave Parker wrote:
Quote: It is well-known that the geometric series 1+a+a^2+a^3... for |a|<1
sums
to 1/(1-a).
The proof of this follows from the following fact applied to the
partial sum:
lim a^n = 0 if |a|<1.
n->oo
Everybody considers this last statement "obvious", but I have never
seen an explicit proof of it.
By applying the definition of limit, the only proof I could find
requires the use of logarithms. (Given an epsilon, you have to find an
n s.t. a^n < epsilon, etc. Logs enter the picture naturally.)
Here is the problem. In most analysis books the development of
logarithms, exponentials, etc. depend on various facts about series,
which when you go through the layers of convergence theorems eventually
end up depending on the convergence of the geometric series. So there
is a chicken and egg problem.
What is an elementary proof of
lim a^n = 0 if |a|<1.
n->oo
that doesn't depend on logarithms or other higher functions?
Thanks.
Dave
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| Peter L. Montgomery |
| Posted: Sat Feb 12, 2005 5:06 pm |
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In article <1108240007.618379.45440@g14g2000cwa.googlegroups.com>
"Dave Parker" <dave3852@yahoo.com> writes:
Quote: What is an elementary proof of
lim a^n = 0 if |a|<1.
n->oo
that doesn't depend on logarithms or other higher functions?
Consider the case 0 < a < 1. Let x = 1/a - 1. Then x > 0 and
(1/a)^n = (1 + x)^n >= 1 + nx = 1 + n*(1/a - 1) = (a + n - n*a)/a
Invert to get
0 <= a^n <= a/(a + n - n*a)
You can prove this directly (i.e., without introducing x)
by induction on n when 0 <= a < 1. Use the sandwich theorem.
--
The Weapons of Mass Destruction are tsunamis.
pmontgom@cwi.nl Microsoft Research and CWI Home: Bellevue, WA |
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| G. A. Edgar |
| Posted: Sat Feb 12, 2005 6:04 pm |
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In article <1108240007.618379.45440@g14g2000cwa.googlegroups.com>, Dave
Parker <dave3852@yahoo.com> wrote:
Quote: It is well-known that the geometric series 1+a+a^2+a^3... for |a|<1
sums
to 1/(1-a).
The proof of this follows from the following fact applied to the
partial sum:
lim a^n = 0 if |a|<1.
n->oo
Everybody considers this last statement "obvious", but I have never
seen an explicit proof of it.
By applying the definition of limit, the only proof I could find
requires the use of logarithms. (Given an epsilon, you have to find an
n s.t. a^n < epsilon, etc. Logs enter the picture naturally.)
Here is the problem. In most analysis books the development of
logarithms, exponentials, etc. depend on various facts about series,
which when you go through the layers of convergence theorems eventually
end up depending on the convergence of the geometric series. So there
is a chicken and egg problem.
What is an elementary proof of
lim a^n = 0 if |a|<1.
n->oo
that doesn't depend on logarithms or other higher functions?
Thanks.
Dave
Many (most?) advanced calculus texts do this with Bernoulli's
inequality.
Bernoulli's Inequality is:
(1+x)^n >= 1+n*x, where x >= -1 and n is a nonnegative integer.
Proved by induction on n. [In fact it is true for
nonnegative *real* n, but at the beginning of such a text
the definition of irrational exponents is not yet known.]
Now from n*x -> infinity as n -> infinity if x>0, we may use
Bernoulli to conclude (1+x)^n -> infinity if x > 0; equivalently
a^n -> infinity if a>1. And from this
we can get a^n -> 0 if 0<a<1 by taking reciprocal.
So ... Do you know how to prove n*x -> infinity if x>0 ???
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/ |
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| Dave Parker |
| Posted: Sun Feb 13, 2005 10:02 am |
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Dave Parker wrote:
Quote: What is an elementary proof of
lim a^n = 0 if |a|<1
n->oo
To everyone who responded: you have answered my question to my
satisfaction, in a variety of interesting and insightful ways. I could
not have asked for a better response, and it exemplifies sci.math at
its
finest. (I'll thank all 5 responders - David Kastrup, Stuart
Newberger,
Johan Mebius, Peter Montgomery, G. A. Edgar - collectively here so as
not to clutter up your busy day with numerous individual posts!)
Dave |
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