Re: Cauchy condensation test handout
master1729
> The purpose of this post is to archive an old
> handout
> of mine on the Cauchy condensation test. The handout
> was
> originally written in January 1998 (I believe I had
> earlier
> prepared a handwritten version, in Jan. or Feb.
> 1997)
> and the audience consisted of the students in a high
> school calculus 3 course that I taught several times
> in
> the late 1990s. The course covered (at a level
> somewhat
> higher than in standard college level calculus
> courses)
> improper integrals, sequences and series, parametric
> equations, some complex variables and other
> miscellaneous
> topics (e.g. an introduction to Fourier series
> expansions),
> partial differentiation, and the beginnings of
> multiple
> integration.
>
> Dave L. Renfro
generalizations might be intresting.
tommy condensation test :)
0 < a_n < a_n+1
a0 + a1 + a2 + a3 + ... < oo <=> 2 a0 + 2 a2 + 2 a4 + ... < oo
x)
in general for rising g(n) E N and g'(n) > 1
sum g(n)a_g(n) < oo <-> sum a_n < oo
e.g.
sum T(n)a_T(n) < oo <-> sum a_n < oo
with T(n) = triangular number.
but probably the cases a_n = oo <=> sum ??? = oo
are the intresting ones.
im convinced earlier work has been done on such matters.
btw
(under trivial conditions)
you can use zeta-expansion of a series :
sum f(n) = sum g(1/n) = sum a0 + a1 1/n + a2 1/n^2 + ...
= a0 + a1 zeta(1) + a2 zeta(2) + ...
and prove almost the entire case ( again under trivial conditions ) with the harmonic sum.
but that is more complicated then neccassary.
also , i suppose you ( dave l renfro ) is already aware of the above.
and thus in a sense my post is more aimed at others.
(not at musatov)
i suspect the handout occured during taylor series radius of convergeance lectures because of the geometric trend.
i think there might be potential in applying multiple combinations of condensation and anti-condensation to find properties - closed forms - boundaries of double/triple series or double/triple products.
is there a non-trivial relation to special functions worth mentioning ?
regards
tommy1729
Tim Norfolk
>
> - Show quoted text -
You might try some of the results on summability theory.
Achava Nakhash, the Loving Snake
>
> 0 < a_n < a_n+1
>
> a0 + a1 + a2 + a3 + ... < oo <=> 2 a0 + 2 a2 + 2 a4 + ... < oo
>
Tommy, my person. If 0 < a_n < a_(n+1), then the series necessarily
diverges since the n'th term is never less than a_1 and hence is
bounded away from 0. Now if, instead, you meant
a_0 > a_1 > a_2 > a_3 > .. > 0
then your generalizatation is trivially true, if you meant, as I
suspect, that the subscripts are all the even numbers rather than the
powers of 2 as I though when I first saw your statement in connection
with the word condenation. Not much condensation going on here. Just
two trivial uses of the comparison test. Why bother to publicize such
a thing? We can all figure out the obvious for ourselves without
effort.
As for results related to the Cauchy Condensation Theorem, and so much
more, Konrad Knopp's book whose title is "Theory and Applcation of
Infinite Series" is an absolutely wonderful and very readable
reference.
Regards,
Achava
W^3
<787955160.5424.125728...@gallium.mathforum.org>,
master1729 <tomm...@gmail.com> wrote:
That's false: Take g(n) = 2n^2 and a_n = 1/n^(3/2). Then sum a_n < oo,
but sum g(n)a_g(n) = sum 2n^2/(2n^2)^(3/2) = oo.
> e.g.
>
> sum T(n)a_T(n) < oo <-> sum a_n < oo
>
> with T(n) = triangular number.
False too, with the same example.
master1729
yes.
in fact i wanted to mention it today.
on the other hand , we can generalize until we reach the entire one variable calculus !
for instance *
integral f(x)dx < oo <=> integral g'(x) f(g(x)) dx < oo
under trivial conditions.
or **
sum f(n) < oo <=> [g(n) - g(n-1)] f(g(n)) < oo
( * / ** note the use of a kind of substitute and derivative/difference )
** for instance sum a_n < oo <=> sum (2^n - 2^n-1) a_2^n
=> sum 2^n-1 a_2^n < oo
which leads to y < oo <=> 2y < oo =>
sum a_n < oo <=> sum 2^n * a_2^n < oo
which was the original cauchy condensation ( QED btw )
etc etc
another intresting idea might be to use this in an attempt to prove erdos conjecture.
regards
the master
tommy1729
master1729
> > tommy condensation test :)
> >
> > 0 < a_n < a_n+1
> >
> > a0 + a1 + a2 + a3 + ... < oo <=> 2 a0 + 2 a2 + 2 a4
> + ... < oo
> >
> <Large Snip>
>
> Tommy, my person. If 0 < a_n < a_(n+1), then the
> series necessarily
> diverges since the n'th term is never less than a_1
> and hence is
> bounded away from 0. Now if, instead, you meant
>
> a_0 > a_1 > a_2 > a_3 > .. > 0
yes of course , sorry typo.
>
> then your generalizatation is trivially true, if you
> meant, as I
> suspect, that the subscripts are all the even numbers
> rather than the
> powers of 2 as I though when I first saw your
> statement in connection
> with the word condenation. Not much condensation
> going on here. Just
> two trivial uses of the comparison test. Why bother
> to publicize such
> a thing? We can all figure out the obvious for
> ourselves without
> effort.
yes of course , thats why i put smiling faces in the text.
however you did snip the more intresting part of the post.
not that i can teach you alot or anything ...
maybe others ...
>
> As for results related to the Cauchy Condensation
> Theorem, and so much
> more, Konrad Knopp's book whose title is "Theory and
> Applcation of
> Infinite Series" is an absolutely wonderful and very
> readable
> reference.
>
> Regards,
> Achava
as i replied to tim , we can generalize it until we reach the entire one variable calculus !!
see my reply to tim.
regards
tommy1729
master1729
> In article
> <787955160.5424.125728...@gallium.ma
i mentioned " under trivial conditions " and " in general "
i probably should have said " usually " or something instead of in general.
anyways , i did not intend to claim " always ".
all the replies are " under trivial conditions ".
regards
tommy1729
David C. Ullrich
wrote:
That doesn't make any sense at all - it's comparing the
terms a_j for j between g(n-1) and g(n) to g(n) a_g(n).
What makes sense as a generalization is this:
sum (g(n) - g(n-1)) a_g(n) < infinity <-> sum a_n < infinity.
Let us know when you've proved that or given a counterexample.
>e.g.
>
>sum T(n)a_T(n) < oo <-> sum a_n < oo
>
>with T(n) = triangular number.
>
>
>but probably the cases a_n = oo <=> sum ??? = oo
>
>are the intresting ones.
>
>
>im convinced earlier work has been done on such matters.
>
>
>btw
>
>(under trivial conditions)
>you can use zeta-expansion of a series :
>
>sum f(n) = sum g(1/n) = sum a0 + a1 1/n + a2 1/n^2 + ...
>
>= a0 + a1 zeta(1) + a2 zeta(2) + ...
>
>and prove almost the entire case ( again under trivial conditions ) with the harmonic sum.
>
>but that is more complicated then neccassary.
>
>
>also , i suppose you ( dave l renfro ) is already aware of the above.
>
>and thus in a sense my post is more aimed at others.
>(not at musatov)
>
>i suspect the handout occured during taylor series radius of convergeance lectures because of the geometric trend.
>
>i think there might be potential in applying multiple combinations of condensation and anti-condensation to find properties - closed forms - boundaries of double/triple series or double/triple products.
>
>is there a non-trivial relation to special functions worth mentioning ?
>
>regards
>
>tommy1729
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
Dave L. Renfro
> As for results related to the Cauchy Condensation Theorem,
> and so much more, Konrad Knopp's book whose title is "Theory
> and Applcation of Infinite Series" is an absolutely wonderful
> and very readable reference.
Also very useful are Thomas Bromwich's "Infinite Series",
which I have a Chelsea Publishing Company reprint of,
and Volume 2 of Hobson's "The theory of Functions of a
Real Variable and the Theory of Fourier's Series", which
I have an old Dover Publications reprint of.
I've noticed that when I post attachments via Math Forum,
my post (even without the attachment) doesn't make it to
google, and I'm guessing it doesn't make it to most (all?)
news servers as well. The reason I posted the Cauchy
condensation handout was so that I could cite the post
URL for some people in another group, where the topic
recently came up in a discussion group for U.S. high
school calculus teachers. Anyway, since this thread
and my posted comments have now made it to google via
tommy1729, I thought I'd give the URL where the 2-page
handout (.pdf file) can be found, in case anyone is
interested in it. [The handout was written for -- and
actually covered in class, along with many other topics
rarely covered in a typical calculus class -- students
in a 3rd semester high school calculus course whose enrollment
consisted almost entirely of students who later attended
colleges such as Caltech, MIT, Rice Univ., etc. Several
now have Ph.D.'s in various science fields, one (that I
know of) in math.]
http://mathforum.org/kb/message.jspa?messageID=6885570
Dave L. Renfro
master1729
lol.
let us know ?
i posted , just like you :
> sum (g(n) - g(n-1)) a_g(n) < infinity <-> sum a_n <
> < infinity.
in a reply to tim.
and i was aware of counterexamples of my first post.
see my other reply also today.
thus basicly : today.
in fact , before you posted !!
soo , did you overlook , or did you steal my content ?
regards
tommy1729
David C. Ullrich
wrote:
Huh? You claimed it worked for the triangular numbers.
It doesn't.
Exactly _what_ "trivial conditions" suffice?
(Hint: You simply got it wrong. The version in my
other post is what's true under certain conditions,
which one might or might not regard as trivial.)
>i probably should have said " usually " or something instead of in general.
>
>anyways , i did not intend to claim " always ".
>
>all the replies are " under trivial conditions ".
>
>regards
>
>tommy1729
David C. Ullrich
David C. Ullrich
wrote:
Your reply to Tim had not appeared when I made my
post.
So you posted a formula that actually _is_ correct
under some conditions.
But you haven't said anything at all until you tell
us what those conditions _are_. The "correct"
version is not always true either. Let us know
when you can actually state something about this
that's actually _true_. _What_ "trivial conditions"
do you have in mind?
master1729
case 1)
0 < a_n+1 < a_n < oo
0 < 1/g_n+1 < 1/g_n < oo
g_n E N
sums start at index n = 0 and g_-1 = 0
sum a_n < oo => sum (g_n - g_n-1) * a_(g_n) < oo
case 2)
0 < a_n+1 < a_n < oo
0 < 1/g_n+1 < 1/g_n < oo
g_n E N
sums start at index n = 0 and g_-1 = 0
lim g_n+1 / g_n = 0
sum a_n < oo <= sum (g_n - g_n-1) * a_(g_n) < oo
since case 1 is a subcondition of case 2 we can even say :
0 < a_n+1 < a_n < oo
0 < 1/g_n+1 < 1/g_n < oo
g_n E N
sums start at index n = 0 and g_-1 = 0
lim g_n+1 / g_n = 0
sum a_n < oo <=> sum (g_n - g_n-1) * a_(g_n) < oo
note however that it may be possible to transform a 'g' that does not satisfy lim g_n+1 / g_n = 0 into another 'G' that does satisfy lim g_N+1 / g_N = 0 ( and the other conditions still hold ( case1 conditions ) )
for instance :
if lim g_N+1 / g_N = 0 and
(g_n - g_n-1) * a_(g_n) <= (G_n - G_n-1) * a_(G_n)
then
sum a_n < oo <=> sum (g_n - g_n-1) * a_(g_n) < oo
still follows.
furthermore it must be noted that 'n' can be replaced by 'p' ; 'p' -> the primes. ( ignoring a_0 and a_1 and thus starting to sum at p = 2 ofcourse and replacing the condition g_-1 = 0 by g_1 = 0 )
case 3
if we have the case 2 conditions and
sum a_n < oo OR sum (g_n - g_n-1) * a_(g_n) < oo
is false then
sum a_n = oo <=> sum (g_n - g_n-1) * a_(g_n) = oo
must follow by the pigeonhole principle.
QED
( note : many summability methods or there failure to converge satisfy similar conditions however more complicated and to long to post here for a man with little time ( me ) )
the master
tommy1729
my reply is a bit late , i had to play a chess match.
( i won btw )
" the best move is pure fiction " alfred brinkmann and tommy1729
master1729
( correction )
THE CONDITIONS :
case 1)
0 < a_n+1 < a_n < oo
0 < 1/g_n+1 < 1/g_n < oo
g_n E N
sums start at index n = 0 and g_-1 = 0
sum a_n < oo => sum (g_n - g_n-1) * a_(g_n) < oo
case 2)
0 < a_n+1 < a_n < oo
0 < 1/g_n+1 < 1/g_n < oo
g_n E N
sums start at index n = 0 and g_-1 = 0
lim g_n+1 / g_n < oo
sum a_n < oo <= sum (g_n - g_n-1) * a_(g_n) < oo
since case 1 is a subcondition of case 2 we can even
say :
0 < a_n+1 < a_n < oo
0 < 1/g_n+1 < 1/g_n < oo
g_n E N
sums start at index n = 0 and g_-1 = 0
lim g_n+1 / g_n < oo
sum a_n < oo <=> sum (g_n - g_n-1) * a_(g_n) < oo
note however that it may be possible to transform a
'g' that does not satisfy lim g_n+1 / g_n < oo into
another 'G' that does satisfy lim g_N+1 / g_N < oo (
and the other conditions still hold ( case1
conditions ) )
for instance :
if lim g_N+1 / g_N < oo and
then
still follows.
case 3
QED
the master
tommy1729
( correction )
master1729
and ullrich got silent ...
master1729
master1729
no wait , that has occured before :)
master1729
cant find a mistake ?
cant admit i am correct ?