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Alternating series question
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Ray Vickson
Apr 21, 2013, 8:17:53 PM4/21/13
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Suppose we have an infinite series of the form sum{(-1)^n * a_n} where all a_n > 0 and limit a_n = 0 as n --> infinity. However, we do NOT assume the a_n are monotone decreasing in n; we just assume they --> 0. Can we still say that the series is convergent?
gus gassmann
Apr 21, 2013, 8:31:55 PM4/21/13
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On 21/04/2013 9:17 PM, Ray Vickson wrote:
> Suppose we have an infinite series of the form sum{(-1)^n * a_n} where all a_n > 0 and limit a_n = 0 as n --> infinity. However, we do NOT assume the a_n are monotone decreasing in n; we just assume they --> 0. Can we still say that the series is convergent?
>
Yes, because the a_n must be bounded.
> Suppose we have an infinite series of the form sum{(-1)^n * a_n} where all a_n > 0 and limit a_n = 0 as n --> infinity. However, we do NOT assume the a_n are monotone decreasing in n; we just assume they --> 0. Can we still say that the series is convergent?
>
Mike Terry
Apr 21, 2013, 8:41:30 PM4/21/13
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"Ray Vickson" <RGVi...@shaw.ca> wrote in message
news:9d4e47cd-e84a-4af8...@googlegroups.com...
1/2 - 1/1 + 1/4 - 1/2 + 1/8 - 1/3 + 1/16 - 1/4 + 1/32 - 1/5 + ...
is divergent, but meets your criteria. (I.e. interleaving the terms of a
convergent geometric series, with the terms of the divergent harmonic
series)
Mike.
news:9d4e47cd-e84a-4af8...@googlegroups.com...
is divergent, but meets your criteria. (I.e. interleaving the terms of a
convergent geometric series, with the terms of the divergent harmonic
series)
Mike.
gus gassmann
Apr 21, 2013, 8:48:18 PM4/21/13
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Ray Vickson
Apr 23, 2013, 4:02:35 AM4/23/13
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On Sunday, April 21, 2013 5:17:53 PM UTC-7, Ray Vickson wrote:
> Suppose we have an infinite series of the form sum{(-1)^n * a_n} where all a_n > 0 and limit a_n = 0 as n --> infinity. However, we do NOT assume the a_n are monotone decreasing in n; we just assume they --> 0. Can we still say that the series is convergent?
Thanks to Mike Terry for a definitive answer.
> Suppose we have an infinite series of the form sum{(-1)^n * a_n} where all a_n > 0 and limit a_n = 0 as n --> infinity. However, we do NOT assume the a_n are monotone decreasing in n; we just assume they --> 0. Can we still say that the series is convergent?
1treePetrifiedForestLane
Apr 24, 2013, 1:59:32 PM4/24/13
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Brun's constant is not transcendental
(the sum of the reciprocals of the twin primes,
including one fifth, twice-summed.
(the sum of the reciprocals of the twin primes,
including one fifth, twice-summed.
Robin Chapman
Apr 25, 2013, 4:40:48 AM4/25/13
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On 24/04/2013 18:59, 1treePetrifiedForestLane wrote:
> Brun's constant is not transcendental
proof?
> Brun's constant is not transcendental
1treePetrifiedForestLane
Apr 26, 2013, 12:14:51 AM4/26/13
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it wasn't published?... I found that
a couple of guys on this forum had found
a closed form, that I knew was good to "all known
(heuristically) digits," but had no construction;
I had found this, many years ago,
while I was perusing Ribenboim's _Prime Number Records_,
and it is a simple geometrical construction.
a couple of guys on this forum had found
a closed form, that I knew was good to "all known
(heuristically) digits," but had no construction;
I had found this, many years ago,
while I was perusing Ribenboim's _Prime Number Records_,
and it is a simple geometrical construction.
Robin Chapman
Apr 26, 2013, 5:02:04 AM4/26/13
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On 26/04/2013 05:14, 1treePetrifiedForestLane top-replied:
Can you give a reference? or will you continue wittering?
1treePetrifiedForestLane
Apr 26, 2013, 2:30:12 PM4/26/13
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it is the second-root of the sum of phi to the second power
and one (sndrt(((sndrt(5) + 1)/2)^2 +1)),
mod parentheticals.
and one (sndrt(((sndrt(5) + 1)/2)^2 +1)),
mod parentheticals.
1treePetrifiedForestLane
Apr 26, 2013, 8:20:25 PM4/26/13
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or, of course,
the secondroot of the sum of phi plus two, but
that was not the construction;
has nothing to do with The Regular Tetragon,
as far as I care.
the secondroot of the sum of phi plus two, but
that was not the construction;
has nothing to do with The Regular Tetragon,
as far as I care.
1treePetrifiedForestLane
Apr 27, 2013, 6:42:40 PM4/27/13
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it'd be nice to have a construction for this,
other form, as well. anyway, let's see, how long,
it takes for someone to find "the" construction,
which I had used to input into the calculator,
whence I recognized the number in Ribbenboim.
other form, as well. anyway, let's see, how long,
it takes for someone to find "the" construction,
which I had used to input into the calculator,
whence I recognized the number in Ribbenboim.
William Elliot
Apr 27, 2013, 9:27:23 PM4/27/13
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(sndrt(((sndrt(5) + 1)/2)^2 +1))
sqr(((1 + sqr 5)/2)^2 + 1) = sqr((1 + 2.sqr 5 + 5)/4 + 1)
= sqr((10 + 2.sqr 5)/4) = 1/2 * sqr(10 + 2.sqr 5)
sqr(((1 + sqr 5)/2)^2 + 1) = sqr((1 + 2.sqr 5 + 5)/4 + 1)
= sqr((10 + 2.sqr 5)/4) = 1/2 * sqr(10 + 2.sqr 5)
Robin Chapman
Apr 29, 2013, 4:32:05 AM4/29/13
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>>>> Brun's constant is not transcendental
>>>
>>> proof?
>
> Can you give a reference? or will you continue wittering?
>>>
>>> proof?
>
> Can you give a reference? or will you continue wittering?
> it is the second-root of the sum of phi to the second power
> and one (sndrt(((sndrt(5) + 1)/2)^2 +1)),
> mod parentheticals.
No proof.
> and one (sndrt(((sndrt(5) + 1)/2)^2 +1)),
> mod parentheticals.
No reference.
More wittering.
Some golden section mysticism.
1treePetrifiedForestLane
Apr 29, 2013, 11:43:55 PM4/29/13
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it was simply a construction,
which happened to be good to as far as I could take it
(on some online calculator, as I recall)
to the "heuristic" value in Ribbenboim. in other words,
it was just lucky that I was serious student
o'Bucky Fuller ... unlike most followers, because
they believe Bucky's say-so, that math was unnecessary,
even though he could do spherical trig,
as the commander of a Naval vessel before radio.
do you find it surprising, that phi is involved?...
I wonder if the other guy has not already gotten a recurrence
relation to generate the twin primes, akin to Lucas sequences.
yes, I'm sure it is the correct formula, although
I would now put it somehwat differently,
as an explicit "constitutional equation
for Big Phi or Brun's constant," or how ever it should be said.
> Some golden section mysticism.
which happened to be good to as far as I could take it
(on some online calculator, as I recall)
to the "heuristic" value in Ribbenboim. in other words,
it was just lucky that I was serious student
o'Bucky Fuller ... unlike most followers, because
they believe Bucky's say-so, that math was unnecessary,
even though he could do spherical trig,
as the commander of a Naval vessel before radio.
do you find it surprising, that phi is involved?...
I wonder if the other guy has not already gotten a recurrence
relation to generate the twin primes, akin to Lucas sequences.
yes, I'm sure it is the correct formula, although
I would now put it somehwat differently,
as an explicit "constitutional equation
for Big Phi or Brun's constant," or how ever it should be said.
> Some golden section mysticism.
1treePetrifiedForestLane
Apr 29, 2013, 11:55:48 PM4/29/13
to
I am pedagogical about "secondpowering & secondrooting,"
partly because of Bucky, but I took it further
to the obvious simplciity of using the unit-diameter sphere
as the proper standard of mensuration,
whereas Bucky was justifiably all on
about the God-am tetrahedron. actually,
there sphere & tetrahedron/tetraasteron are really the same thing,
in plenty of ways.
cf, the "lunes proof" of the pythagorean theorem,
clearly the original proof -- even if Hippocrates (not, though,
the medical one) came after Pythagoras -- and
extend it to the two spatial analogs.
partly because of Bucky, but I took it further
to the obvious simplciity of using the unit-diameter sphere
as the proper standard of mensuration,
whereas Bucky was justifiably all on
about the God-am tetrahedron. actually,
there sphere & tetrahedron/tetraasteron are really the same thing,
in plenty of ways.
cf, the "lunes proof" of the pythagorean theorem,
clearly the original proof -- even if Hippocrates (not, though,
the medical one) came after Pythagoras -- and
extend it to the two spatial analogs.
Robin Chapman
Apr 30, 2013, 4:20:24 AM4/30/13
to
On 30/04/2013 04:43, 1treePetrifiedForestLane wrote:
<snip wittering>
<snip more wittering>
<snip wittering>
>
> do you find it surprising, that phi is involved?...
golden section mysticism comes as no surprise on sci.mayj
> do you find it surprising, that phi is involved?...
<snip more wittering>
1treePetrifiedForestLane
Apr 30, 2013, 9:24:12 PM4/30/13
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you cannot escape phi with the icosahedron,
that is all there is to it, and this is nothing
but a simple observation.
I assume that this will prove to be a good way
to develop a new second-rooting method, even if
there is no direct application to twin-primes.
> golden section mysticism comes as no surprise on sci.mayj
so, what is the constitutive equation for Big Phi?
that is all there is to it, and this is nothing
but a simple observation.
I assume that this will prove to be a good way
to develop a new second-rooting method, even if
there is no direct application to twin-primes.
> golden section mysticism comes as no surprise on sci.mayj
Robin Chapman
May 1, 2013, 4:27:49 AM5/1/13
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On 01/05/2013 02:24, 1treePetrifiedForestLane wrote:
<golden section wittering deleted>
> I assume that this will prove to be a good way
> to develop a new second-rooting method, even if
> there is no direct application to twin-primes.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
So, despite your earlier comments, you admit that
your alleged formula for Brun's constant was bogus.
>> golden section mysticism comes as no surprise on sci.math
So, any reference or proof for your formula for Brun's constant?
<golden section wittering deleted>
> I assume that this will prove to be a good way
> to develop a new second-rooting method, even if
> there is no direct application to twin-primes.
So, despite your earlier comments, you admit that
your alleged formula for Brun's constant was bogus.
>> golden section mysticism comes as no surprise on sci.math
>
> so, what is the constitutive equation for Big Phi?
WTF are you talking about?
> so, what is the constitutive equation for Big Phi?
So, any reference or proof for your formula for Brun's constant?
1treePetrifiedForestLane
May 1, 2013, 3:08:33 PM5/1/13
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what, do you say, is bogus about it?
you do not seem to be conversant with Fibonacci or
Lucas sequences, so you have nothing to say
in defense of your thesis -- what ever it is,
que sera, sera.
> So, despite your earlier comments, you admit that
> your alleged formula for Brun's constant was bogus.
I am tentatively callin the secondroot
of the sum of two and phi, Big Phi; maybe,
i'll be able to extend it to a little scheme,
phi, phi_big, phi_bigger, phi_biggest;
phinitessimals & phiinfinities?
you do not seem to be conversant with Fibonacci or
Lucas sequences, so you have nothing to say
in defense of your thesis -- what ever it is,
que sera, sera.
> So, despite your earlier comments, you admit that
> your alleged formula for Brun's constant was bogus.
of the sum of two and phi, Big Phi; maybe,
i'll be able to extend it to a little scheme,
phi, phi_big, phi_bigger, phi_biggest;
phinitessimals & phiinfinities?
1treePetrifiedForestLane
May 1, 2013, 5:44:34 PM5/1/13
to
no plateaux; eh?
Robin Chapman
May 2, 2013, 4:39:25 AM5/2/13
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On 01/05/2013 20:08, 1treePetrifiedForestLane wrote:
>
>> So, despite your earlier comments, you admit that
>> your alleged formula for Brun's constant was bogus.
>
> I am tentatively callin the secondroot
> of the sum of two and phi, Big Phi; maybe,
> i'll be able to extend it to a little scheme,
> phi, phi_big, phi_bigger, phi_biggest;
> phinitessimals & phiinfinities?
No proof of any relevance to Brun's constant, but
>
>> So, despite your earlier comments, you admit that
>> your alleged formula for Brun's constant was bogus.
>
> I am tentatively callin the secondroot
> of the sum of two and phi, Big Phi; maybe,
> i'll be able to extend it to a little scheme,
> phi, phi_big, phi_bigger, phi_biggest;
> phinitessimals & phiinfinities?
a plethora of still-born phi-based neologisms.
It's all bullphit, isn't it?
Robin Chapman
May 2, 2013, 4:40:23 AM5/2/13
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1treePetrifiedForestLane
May 2, 2013, 10:17:38 PM5/2/13
to
the formula, compared to the heuristical value
of Brun's constant, is more than adequate;
I have simply been putting it as,
"Brun's constant is not transcendental;
assume that it is & derive a contradiction."
> It's all bullphit, isn't it?
you appear to be doing that!
of Brun's constant, is more than adequate;
I have simply been putting it as,
"Brun's constant is not transcendental;
assume that it is & derive a contradiction."
> It's all bullphit, isn't it?
1treePetrifiedForestLane
May 2, 2013, 10:19:33 PM5/2/13
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diverges to minus infinity?
Robin Chapman
May 3, 2013, 4:17:55 AM5/3/13
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On 03/05/2013 03:17, 1treePetrifiedForestLane wrote:
> the formula, compared to the heuristical value
> of Brun's constant, is more than adequate;
Does that mean true, or not?
> the formula, compared to the heuristical value
> of Brun's constant, is more than adequate;
> I have simply been putting it as,
> "Brun's constant is not transcendental;
> assume that it is & derive a contradiction."
or any reference .... Are you sure you didn't
dream it?
dull...@sprynet.com
May 3, 2013, 10:57:46 AM5/3/13
to
Brun's constant is obviously rational, being the
sum of rationals. Maybe you're stuck on showing
that a rational number cannot be transcendental?
I don't recall exactly how that's proved...
Heh.
1treePetrifiedForestLane
May 5, 2013, 3:11:16 PM5/5/13
to
not my field, but I'm sure that you can find many easy examples
of infinite sums that converge to irrational values,
viz Liebniz determination of pi fourths
using whats-is-names arctangent thing.
we know you can, Dullr!
> Brun'sconstant is obviously rational, being the sum of rationals.
of infinite sums that converge to irrational values,
viz Liebniz determination of pi fourths
using whats-is-names arctangent thing.
we know you can, Dullr!
> Brun'sconstant is obviously rational, being the sum of rationals.
1treePetrifiedForestLane
May 6, 2013, 3:26:52 PM5/6/13
to
here's a cool problem; I changed two of the variables.
K/(fourthroot(vv/2)) = (fourthroot(vv/2)/M
K/(fourthroot(vv/2)) = (fourthroot(vv/2)/M
Robin Chapman
May 7, 2013, 6:01:41 AM5/7/13
to
On 06/05/2013 20:26, 1treePetrifiedForestLane wrote:
> here's a cool problem; I changed two of the variables.
>
> K/(fourthroot(vv/2)) = (fourthroot(vv/2)/M
So much for the Millennium Prize problems!
> here's a cool problem; I changed two of the variables.
>
> K/(fourthroot(vv/2)) = (fourthroot(vv/2)/M
1treePetrifiedForestLane
May 7, 2013, 6:11:16 PM5/7/13
to
such perseverence. unforunately,
they did not include the characterization of the Fermat primes.
they did not include the characterization of the Fermat primes.
1treePetrifiedForestLane
May 9, 2013, 12:52:38 AM5/9/13
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because, the MP cmte. *knew* that I was working on it?... well,
there's always the Riemann thing!
there's always the Riemann thing!
Spac...@hotmail.com
Jun 26, 2013, 10:49:00 PM6/26/13
to
hint: use one of the 3d pythagorean theorems; also,
may be it *was* the medical Hippocrates,
who found the Lunes Proof; was it?
also, the 13 books of Euclid are horribly encyclopedic;
one can easily begin with -- a bit of maturity --
the 14th book, by Hypsicles,
which is just solid geometry;
beautifully elementary theorems
> the secondroot of the sum of phi plus two, but
> that was not the construction;
may be it *was* the medical Hippocrates,
who found the Lunes Proof; was it?
also, the 13 books of Euclid are horribly encyclopedic;
one can easily begin with -- a bit of maturity --
the 14th book, by Hypsicles,
which is just solid geometry;
beautifully elementary theorems
> the secondroot of the sum of phi plus two, but
> that was not the construction;
Spac...@hotmail.com
Jun 26, 2013, 10:55:44 PM6/26/13
to
all so good. now,
let us have a "lunes" proof of either
of the 3d pythagoreans; thank you
let us have a "lunes" proof of either
of the 3d pythagoreans; thank you
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