HDTenEightyP/NetNews huggingface dataset:
From: dfs@doe.carleton.ca (David F. Skoll)
Subject: Re: The negative Harmonic series
Organization: Dept. of Electronics, Carleton University
Date: Mon, 13 Apr 1992 16:40:47 GMT

In fact, the alternating harmonic series can be "arranged" to sum
to anything you like, or even to diverge.

Pick a number Z to which you'd like the series to "converge."  Sum all
the positive terms until the result is a bit higher than Z (For
negative Z, pick negative terms until the result is a bit lower than
Z):

1 + 1/3 + 1/5 + 1/7 + ... = Z + epsilon.

Now mix in some negative terms, until it goes below Z (Again, for negative
Z, reverse the sign and the sense of the inequality):

1 + 1/3 + 1/5 + 1/7 + ... - 1/2 - 1/4 - ... = Z - epsilon2.

Then add in some remaining positive terms and so on.  Your series will contain
all the terms, and will sum arbitrarily close to Z.

To make the series "diverge", use positive terms until the sum is 10, then
negative terms until it is -100, then positive until it is 1000, and so on.
You'll get an unbounded, oscillating sequence of sums.

--
David F. Skoll.
From: carabalo@phoenix.Princeton.EDU (David G. Caraballo)
Newsgroups: rec.puzzles
Subject: Re: A simple series question...
Date: 6 Apr 92 21:09:48 GMT
Article-I.D.: Princeto.1992Apr6.210948.13076
Organization: Princeton University
Originator: news@ernie.Princeton.EDU

In article <1992Apr3.083118.13316@Princeton.EDU> carabalo@phoenix.Princeton.EDU (David G. Caraballo) writes:
>The series 1 - 1/2 + 1/3 - 1/4 + 1/5 + ... + (-1)^n/(n+1) + ...
>clearly converges (but not absolutely, so be careful about rearranging terms!)
>and the sum is well known.  Find a closed form for this sum.  I will post a
>solution if no one else does so fairly soon.  Enjoy.  
>
>David

I received several numerical approximations but no complete solution.  The sum
is simply ln 2.  Equating the terms above with the general form for a Taylor
series expansion gives properties about the various derivatives of the 
function in question, and these lead immediately to the function ln x at x=2.
This general method works for many infinite series.

I had hoped to receive a more elementary solution, one that made no use of
Taylor series, since the series I posted appears to be so elementary.  One
does not need Taylor series, for example, to prove that the sum of the first
n odd integers (positive) is n^2.

David G. Caraballo
From: morgan@spectra.com (Mike Morgan)
Newsgroups: rec.puzzles
Subject: Re: The Foreign Counterfeiter
Date: 11 Apr 92 16:19:40 GMT
Article-I.D.: spectra.1992Apr11.161940.21296
Organization: Spectragraphics Corporation

In article <YdtPlVK00iV1M2Gc8M@andrew.cmu.edu> ss8c+@andrew.cmu.edu (Simon Szykman) writes:
>
>OBFAQ-QUESTION-THAT-EVERYBODY-HATES-SEEING-BUT-SOMEONE-INEVITABLY-
>    POSTS-IT-EVERY-WEEK-ANYWAY:
>
>What is the next number in this series?
>
>1, 11, 21, 1211, 111221, ?

FFFFF  IIIII   V   V   EEEEE      ???
F        I     V   V   E         ?   ?
FFF      I      V V    EEE          ?
F        I      V V    E           ?
F      IIIII     V     EEEEE
                                   ?
-- 
Michael L. Morgan 
morgan@spectra.com
From: tom@ulysses.att.com (Tom Smith)
Newsgroups: rec.puzzles
Subject: Re: A question of addresses
Keywords: A simple puzzle for Wednesday
Date: 9 Apr 92 03:58:30 GMT
Reply-To: tom@ulysses.UUCP (Tom Smith)
Distribution: na
Organization: AT&T Bell Laboratories, Murray Hill

*SPOILER WARNING*


In article <1992Apr8.202556.29597@clipper.ingr.com> kontos@clipper.ingr.com (Thorne Kontos) writes:
>Sally and Sue have a strong desire to date Sam. They all live on the
>same street yet neither Sally or Sue know where Sam lives. The houses
>on this street are numbered 1 to 99.
>
>Sally asks Sam "Is your house number a perfect square?". He answers.
>Then Sally asks "Is is greater than 50?". He answers again.
>
>Sally thinks she now knows the address of Sam's house and decides to
>visit.

	Since Sally thinks that she has enough information,
	I deduce that Sam answered that his house number was
	a perfect square greater than 50.  There are two
	of these {64,81} and Sally must live in one of them in
	order to have decided she knew where Sam lives.

>When she gets there, she finds out she is wrong. This is not
>surprising, considering Sam answered only the second question
>truthfully.

	So Sam's house number is greater than 50, but not
	a perfect square.

>Sue, unaware of Sally's conversation, asks Sam two questions.
>Sue asks "Is your house number a perfect cube?". He answers.
>She then asks "Is it greater than 25?". He answers again.

	Observation: perfect cubes greater than 25 are
	{27, 64}, less than 25 are {1,8}.

>Sue thinks she knows where Sam lives and decides to pay him a visit.
>She too is mistaken as Sam once again answered only the second 
>question truthfully.

	Since Sam's house number is >50, he told Sue that it
	was >25 as well.  Since Sue thought she knew which house
	was his, she must live in either of {27,64}.

>If I tell you that Sam's number is less than Sue's or Sally's,

	Since Sam's number is >50, and Sue's is even bigger, she
	must live in 64.  Assuming Sue and Sally are not roommates
	(although awkward social situations of this kind are not
	without precedent), Sally lives in 81.

>and that the sum of their numbersis a perfect square multiplied
>by two, you should be able to figure out where all three of them
>live.

	Sue + Sally + Sam = 2 p^2        for p an integer
	64  + 81    + Sam = 2 p^2

	Applying the constraint 50 < Sam < 64, looks like Sam=55 (p=10).

	In summary,
		Sam = 55
		Sue = 64
		Sally = 81

	Pretty cool puzzle.  Probably because I could solve it.

Tom Smith.
tom@ulysses.att.com
Newsgroups: rec.puzzles
From: carl@SOL1.GPS.CALTECH.EDU (Carl J Lydick)
Subject: Re: trvia
Reply-To: carl@SOL1.GPS.CALTECH.EDU
Organization: HST Wide Field/Planetary Camera
Date: Sat, 11 Apr 1992 23:01:40 GMT

In article <703024532.1@nwark.FidoNet>, Marc.Costaldi@f2050.n391.z1.FidoNet.Org (Marc Costaldi) writes:
>Got a problem need some help with:
>
>An architect was asked to design a square house with the window in each
>wall 
>having a southern exposure. How was this done?
>
>any ideas?
Spoiler

He built it at the north pole.
--------------------------------------------------------------------------------
Carl J Lydick | INTERnet: CARL@SOL1.GPS.CALTECH.EDU | NSI/HEPnet: SOL1::CARL

Disclaimer:  Hey, I understand VAXes and VMS.  That's what I get paid for.  My
understanding of astronomy is purely at the amateur level (or below).  So
unless what I'm saying is directly related to VAX/VMS, don't hold me or my
organization responsible for it.  If it IS related to VAX/VMS, you can try to
hold me responsible for it, but my organization had nothing to do with it.
From: ss8c+@andrew.cmu.edu (Simon Szykman)
Newsgroups: rec.puzzles
Subject: A Make a Puzzle Puzzle
Date: 14 Apr 92 16:40:05 GMT
Organization: Masters student, Mechanical Engineering, Carnegie Mellon, Pittsburgh, PA


Here's one that's kinda tough if you actually sovle a
puzzle by trying to do it yourself, instead of writing
a program to look up words in a dictionary.

Consider the well known "code"
a = 1
b = 2
c = 3
...

1.   Make a magic word square (i.e. one of those n x n
squares where reading all rows, columns and diagonals are
words) where translating the letters into numbers using the
above code yields a regular magic square (i.e. all rows,
columns and diagonals add up to the same number).

2.   Same as 1, but without repeating words (this is 
considerably tougher).

Clearly, there is no way to do either of these for n = 2, 
and n = 1 has only trivial answers - "a" = "1" and "i" = "9",
so lets start with n > 2.

                            - Simon
Newsgroups: rec.puzzles
From: jwd2@screech.owlnet.rice.edu (Joseph William DeVincentis)
Subject: Re: Primes in sequences...
Organization: Rice University
Date: Tue, 14 Apr 1992 01:25:31 GMT

In article <1992Apr13.163322.7068@Princeton.EDU>, carabalo@phoenix.Princeton.EDU (David G. Caraballo) writes:
> I canceled my post about the sequence 1, 11, 111, 1111, ..., since it contained
> a typo that invalidated the proof.  In fact, I have been unable to construct
> a proof that there are only a finite number of primes in that sequence, and I
> am not entirely sure this is the case.  If a_n is the term with n 1's, then
> a_n is clearly composite if n is composite:  If r>1 divides n, then a_r divides
> a_n; thus, it suffices to check a_n for prime values of n.
> 
> Q1:  What is the least odd prime p for which a_p is prime?  

Maple tells me that a_19 and a_23 are prime. All other a_n for
prime n 3 to 29 are composite.

-- 
\  ,------------------------------------------------------------.  /
 \/     Joseph  De Vincentis            jwd2@owlnet.rice.edu     \/
 /\           Help! My .sig just flipped umop-ap!sdn !           /\
/  `------------------------------------------------------------'  \
From: citroen@binoc.bih.harvard.edu (Micky Citroen)
Newsgroups: rec.puzzles
Subject: Re: The Philosopher's Puzzle...
Date: 10 Apr 92 17:43:36 GMT
Reply-To: citroen@binoc.bih.harvard.edu (Micky Citroen)
Followup-To: rec.puzzles
Organization: Beth Israel Hospital, Harvard Medical School, Boston Mass., USA

In article <1992Apr9.122034@IASTATE.EDU> macster@IASTATE.EDU (Michael C Mccarty) writes:

[..]
>Imagine you are one of 3 philosophers sitting under a tree in the middle of
>a nice meadow.  All of you fall asleep under the tree.  Meanwhile a group of
>birds come to perch in the tree while you are sleeping. The birds then proceed
>to sh*t on all three of your heads.  You all wake up simultaneously and, while
>none of you can feel the sh*t on top of your head, you all burst out laughing
>because you see sh*t on the others heads.Now assuming that none of you noticed
>who the others were looking at when they started laughing,and that none of you
>bother to feel your head (pretty big assumptions,I know, but please, bear with
>me), how can you determine that you have sh*t on your head?
>
>I must note that this isn't really much of a puzzle but it is a good refresher
>course in good, clear, logical thinking (which I think some of you are in dire
>need of).

How about asking one of the other two "Do I have sh*t on my head?". Or
am I missing some hidden restriction here?

Micky.








Newsgroups: rec.puzzles
From: guckes@math.fu-berlin.de (Sven Guckes)
Subject: Re: Math (?) puzzle
Originator: guckes@medusa
Organization: Free University of Berlin, Germany
Date: Thu, 16 Apr 1992 00:06:30 GMT

brent-b@eastern.COM (Brent McNamee) writes:

>Hey all...
>You may have seen this one before, but here goes anyway...
>What are the next two (or three, etc...) numbers in the following series?
>NOTE: it is NOT a mathematical series...
WHAT? Of course it is! You can define it by recursion (or whatever).

>1
>11
>21
>1211
>111221
>312211
>13112221
>1113213211
 31131211131221
 13211311123113112211

Somebody please add this to the FAQ. It pops up every two weeks now.

Sven
From: hamel@primate.wisc.edu (Ray Hamel)
Newsgroups: rec.puzzles
Subject: SWD #4 -- Answer
Date: 8 Apr 92 18:29:12 GMT
Organization: UW-Madison Primate Center


DAEDALUSRUSSELLEBERTSERRATEDACTUALMUSTANG
-----------------------------------------
         11111111112222222222333333333344
12345678901234567890123456789012345678901
 
 
 
   FORWARD
 1- 8     Winged mythological figure
 9-15     Jehovah's Witnesses founder
16-20     Famed film critic
21-28     Like a bread knife
29-34     Tangible
35-41     Classic auto
 
   BACKWARD
41-37     Small insects
36-31     Pronounciation symbol
30-26     Future commissioned officer
25-20     Stop
19-15     Reject authority
14-11     Fewer
10- 5     "Little Mermaid" villainess
 4- 1     Without a charge

Ray Hamel
University of Wisconsin
From: js06+@andrew.cmu.edu (Jason Lee Smith)
Newsgroups: rec.puzzles
Subject: Crpytic Puzzle Solving
Date: 8 Apr 92 15:29:11 GMT
Article-I.D.: andrew.sdsl57K00Vpa9=P1Qj
Organization: Junior, Mechanical Engineering, Carnegie Mellon, Pittsburgh, PA

Does anybody have a guide to; Cryptic puzzle solving that they'd  be
willing to post to the net. I think I need a refresher course, the only
clues I seem to be able to get, lately, are anagrams.


							J
From: uslave@mcl.mcl.ucsb.edu (KJS @ucsb.mcl)
Newsgroups: rec.puzzles
Subject: Re: Monday morning brain energizer April 6th
Keywords: A weekly series in the making?
Date: 7 Apr 92 06:55:20 GMT
Distribution: na

*** B: "E was not first" ***

AARGH!!! I forgot about that.  Well, that means that my stating E as being
#1 was wrong.  He's #2, and A is #1.   

I think I used 'E is #1' as a foundation to work on because I thought it 
read, "E: 'I was second'" instead of what it was, "E: 'B was second'".

So, the actual answer is: A, E, B, C, D.

--
*        *      *     *     *    *   *  *  * * ** ** *** **** ****** ********* 
   KJS @ucsb.mcl (Jesse Chang)               e-mail: uslave@mcl.mcl.ucsb.edu *
   "Traveling thru hyperspace ain't like dusting crops, boy!"    -- Han Solo * 
*         *       *      *     *    *   *  *  * ** ** *** **** ****** ********
From: tvaughan@central.rutgers.edu (Timothy E. Vaughan)
Newsgroups: rec.puzzles
Subject: A Modest Proposal
Date: 14 Apr 92 13:43:08 GMT
Organization: Rutgers Univ., New Brunswick, N.J.


Despite the Subject of this posting, it is not intended to be
sarcastic.

I propose that people who know the answer to a posted puzzle simply
because they heard it before (frequently LONG before) refrain from
posting an answer.  People who genuinely FIGURED OUT a puzzle that was
NEW TO THEM should happily post.  I believe the savings in bandwidth
would be astounding.

Tim
Newsgroups: rec.puzzles
From: james@crc.ricoh.com (James Allen)
Subject: Re: Game Theory
Organization: RICOH California Research Center
Date: Wed, 15 Apr 92 04:28:02 GMT

In <1992Apr14.000319.1@tesla.njit.edu> jpk2581@tesla.njit.edu writes:
]
]...I've got a game I would like analyzed:
	[ see original article for description of game ]

	First note that this game is a highly simplified version of
	_Clue_ by Parker Bros.  That game is played with a gameboard,
	dice, a toy candlestick, etc; but a "better" version of that game
	can be played with no props at all except an ordinary deck
	of cards.  My friends and I played this a few times a few
	decades ago; it seemed relatively interesting as a logic game.
]
]Each player ante's one chip. The players then alternate turns
]having the option on each turn to either put 
]another chip in the pot and ask the opponent if

	If you want a tractable game you may prefer not to increase
	the pot: see below.

]The game is a lot more subtle than it first appears.
]	...if you bluff and your opponent 
]does not flip the card on the next turn then he will
]KNOW you were bluffing (and thus know one of the cards 
]in your hand) when you don't know the value of the
]hidden card.

	No; things are "even worse" than this.  If I know
	what the secret card is I may prefer to keep adding chips
	and asking so that the opponent will also continue to
	sweeten the pot.
]
]Well, anyway the game doesn't seem like it's overly 
]difficult to solve, but it has escaped me....

	Assuming you want a closed-form exact solution, it *does*
	seem rather difficult to me.  I've tried to analyze
	various versions of Poker but I've succeeded only for
	versions much simpler than your game.

	Let me present a *much* simpler game which many rec.puzzlers
	will be able to solve, but you may be surprised how difficult
	the solution is.

	What is the simplest game in the Whist family?  Deal two
	cards to each of two players, leaving 48 cards unexposed
	and out of play.  The dealer names trump; the other player
	leads and "play continues" as in any whist game.  Dealer
	wins V0, V1, or V2 depending on whether he takes 0, 1 or 2
	tricks.  V0 < V1 < V2.  For definiteness you might take
		V0 = -1.0
		V1 = -0.7647
		V2 = +1.0
	(This happens to describe the last round of a popular
	child's whist variant which goes by various names
	including "Seven Up" or "Sargeant Major".)

	The only decisions in this game are (1) what to name as
	trump holding two singletons, and (2) what to lead holding
	a high singleton trump.  Since "V1" is small you should
	declare Hearts as trump holding Spade Ace and Heart Deuce.
	But complete solution involves an additional subtlety.
]
]Would any of you game theoritians care to give this one 
]a try?

	If you don't need an exact solution, may I recommend
	the Genetic Algorithm, eg Robert Axelrod, "The Evolution
	of Strategies in the Iterated Prisoner's Dilemma" in
	_Genetic Algorithms and Simulated Annealing_ by Davis,
	pub 1987 Morgan Kaufman.

James Allen		-- james@crc.ricoh.com
From: dant@logos.wr.tek.com (Dan Tilque)
Newsgroups: rec.puzzles
Subject: Re: Missing subject header
Date: 9 Apr 92 23:12:03 GMT
Organization: Upper Left Coast Liberation Army

Demoel@Fwva.Saic.Com (Ed de Moel) writes:
>news@watson.ibm.com (NNTP News Poster) writes:
>>
>>Prove or disprove that the number of primes in the sequence
>>1 , 11 , 111 ,  1111 , 11111 , ...
>>is finite. 
> 
>Since 111 is divisable by 3 and 1111 is divisable by 11 that shouldn't
>be too hard...

Those with a composite number of digits are divisible by at least one
other number in the list.  However, that still leaves us with an infinite
number of those with a prime number of digits.

It seems to me that I've seen this one posted before, but it doesn't seem
to be in the FAQ index.

---
Dan Tilque	--	dant@logos.WR.TEK.COM
From: awpaeth@taurus.ece.cmu.edu (Alan Paeth)
Newsgroups: rec.puzzles
Subject: Re: primes in 1, 11, 111, etc.
Date: 12 Apr 92 21:34:06 GMT
Organization: Electrical and Computer Engineering, Carnegie Mellon

In <sa4h9INNrt5@agate> davidb@triangle.Berkeley.EDU (David T.  Blackston) sez:
>
>Here are some interesting (?) related questions.  Let a_n be the 
>number consisting of n ones.
>Question 1: Show that a_n is prime only if n is prime.  (Easy)
---

Thanks for a trip down memory lane, David. Your question did interest me.
Twenty years ago this May I wrote an exam for gifted HS math students to see
who might attend a university-run math camp and work on main-frames (ohh, ah!).
The one question interesting enough to remember was this:

   "Prove (non)prime: the decimal number 111....1 having ninty-one digits".

The first "aha!" was that 91 is a very choice number, being <100 yet composite,
as it neither appears in grade school times table, nor reveals its factors by
simple inspection (eg, 3, 5 or 11). This bit of suspician quickly lead to the
factoring. When written out, it has the appearance of binary multiplication, as
only 0's and 1's are involved, with no columns containing more than one unit.
In all, a very good question for a CS major-to-be.

    /Alan Paeth

PS - 23 was a favorite prime then; so (10^23-1)/9 quickly followed "naturally"
:-)
Newsgroups: rec.puzzles
From: carl@SOL1.GPS.CALTECH.EDU (Carl J Lydick)
Subject: Re: The negative Harmonic series
Reply-To: carl@SOL1.GPS.CALTECH.EDU
Organization: HST Wide Field/Planetary Camera
Date: Wed, 15 Apr 1992 00:13:59 GMT

In article <1992Apr11.020521.16082@Cadence.COM>, hemmady@Cadence.COM (Shankar Hemmady) writes:
>
>
>The series that David G. Caraballo refers to is the well-known
>harmonic series with alternately positive and negative terms.
>
>It has puzzled mathematicians for a long time.
>
>It satisfies the rules of convergence, but not those of absolute
>convergence.
>
>So here is a puzzle on that subject.  (David please do not answer this
>one)
>
>	1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 ....
>
>   =	1 - (2/2-1/2) + 1/3 - (2/4-1/4) + 1/5 - (2/6-1/6) + 1/7 - (2/8-1/8) ....
>
>	expanding the brackets and re-grouping ...
>
>   = 	(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...)
>	- (2/2 + 2/4 + 2/6 + 2/8 + ...)
>
>	factorizing ther 2/2, 2/4, etc ...
>
>   = 	(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...)
>     	- (1 + 1/2 + 1/3 + 1/4 + ...)
>
>   =    ZERO!!
>
>
>What went wrong.  I know that there is the Absolute Convergence theorem which
>can find the error in the above statements.  But can you give a 
>simple lay-man explanantion of the fallacy?

Sure.  The next to last statement of the series is infinity minus infinity. 
You can't do that.
--------------------------------------------------------------------------------
Carl J Lydick | INTERnet: CARL@SOL1.GPS.CALTECH.EDU | NSI/HEPnet: SOL1::CARL

Disclaimer:  Hey, I understand VAXes and VMS.  That's what I get paid for.  My
understanding of astronomy is purely at the amateur level (or below).  So
unless what I'm saying is directly related to VAX/VMS, don't hold me or my
organization responsible for it.  If it IS related to VAX/VMS, you can try to
hold me responsible for it, but my organization had nothing to do with it.
From: surya@watson.ibm.com (Surya Mohanty)
Newsgroups: rec.puzzles
Subject: Re: Simple NT problem
Date: 14 Apr 92 19:28:23 GMT
Reply-To: surya@rothko.watson.ibm.com (Surya Mohanty)
Organization: IBM T.J. Watson Research Center
Disclaimer: This posting represents the poster's views, not necessarily those of IBM

In article <1992Apr14.170517.1768@clpd.kodak.com>,
benham@anion.serum.kodak.com (Todd Benham (x37354)) writes:
> In article <1992Apr13.160726.9198@watson.ibm.com>
surya@rothko.watson.ibm.com (Surya Mohanty) writes:
> >If a and b are positivite integers and if 
> >     1+ a^2 + b^2
> >k= --------------
> >       a*b
> >is also an integer then prove that k is equal to 3.
> 
> 
> a=2,b=4 .: k=21/8
> 
> You must be mistaken.
> 
If you read the question carefully you will note that it requires k to
be an integer. k = 21/8 IS NOT AN INTEGER.
From: torbenm@diku.dk (Torben AEgidius Mogensen)
Newsgroups: rec.puzzles
Subject: Re: The negative Harmonic series
Date: 13 Apr 92 15:14:30 GMT
Organization: Department of Computer Science, U of Copenhagen

hemmady@Cadence.COM (Shankar Hemmady) writes:



>The series that David G. Caraballo refers to is the well-known
>harmonic series with alternately positive and negative terms.

>It has puzzled mathematicians for a long time.

>It satisfies the rules of convergence, but not those of absolute
>convergence.

>So here is a puzzle on that subject.  (David please do not answer this
>one)

>	1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 ....

>   =	1 - (2/2-1/2) + 1/3 - (2/4-1/4) + 1/5 - (2/6-1/6) + 1/7 - (2/8-1/8) ....

>	expanding the brackets and re-grouping ...

>   = 	(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...)
>	- (2/2 + 2/4 + 2/6 + 2/8 + ...)

    =   infinity - infinity         !!!!!!!!!!!!!!!!!!!!

>	factorizing ther 2/2, 2/4, etc ...

>   = 	(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...)
>     	- (1 + 1/2 + 1/3 + 1/4 + ...)

>   =    ZERO!!


>What went wrong.  I know that there is the Absolute Convergence theorem which
>can find the error in the above statements.  But can you give a 
>simple lay-man explanantion of the fallacy?

Infinity minus infinity is not well defined.

	Torben Mogensen (torbenm@diku.dk)
From: clong@romulus.rutgers.edu (Chris Long)
Newsgroups: rec.puzzles
Subject: Re: Math Problem
Date: 11 Apr 92 00:35:04 GMT
Organization: Rutgers Univ., New Brunswick, N.J.

> james@crc.ricoh.com (James Allen) writes...

>> sod ( sod ( sod (4444 ^ 4444 ) ) )

We've all seen this one before ... now, what is the largest x for
which we can easily compute sod^3(x)?

-Chris
Newsgroups: rec.puzzles
From: karr@cs.cornell.edu (David Karr)
Subject: Re: The negative Harmonic series
Organization: Cornell Univ. CS Dept, Ithaca NY 14853
Date: Mon, 13 Apr 1992 15:29:28 GMT

In article <1992Apr11.020521.16082@Cadence.COM> hemmady@Cadence.COM (Shankar Hemmady) writes:
>
>	1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 ....
>
[...intermediate steps omitted...]
>
>   = 	(1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...)
>     	- (1 + 1/2 + 1/3 + 1/4 + ...)
>
>   =    ZERO!!
>
>What went wrong.  I know that there is the Absolute Convergence theorem which
>can find the error in the above statements.  But can you give a 
>simple lay-man explanantion of the fallacy?

How about the fact that 1 + 1/2 + 1/3 + ... sums to infinity?  Then you
are subtracting infinity from infinity, which is a well-known fallacy.

-- David Karr
Newsgroups: rec.puzzles