What is the limit of the sequency (1/2)*(3/4)*(5/6)...... when n tends
to the infinite ? The same for the sequence (2/3)*(4/5)....How can I
find the solution?

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Clarification of Question by brandmath-ga on 22 Dec 2005 04:11 PST

I thank both really brilliant comments. Nevertheless, my expectation
was that the answer had the Euler number (e). My question is part of a
very difficult probability problem and the final answer must be
presented in terms of e. Is it possible with this two series or we
only can put the answer in terms of pi?

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Clarification of Question by brandmath-ga on 23 Dec 2005 10:46 PST

In order to clarify the context of my question I will present the
complete problem that origined it. Well, we have an urn with n balls.
All of them are white but one that is black. Its a game played by two.
One of them canīt see the balls. The rule is, first the blind must
take one ball off the urn next the other must take one ball white
(must be white), and so on. Whatever happens, the last ball must be
taked off by the blind. The blind will win if the last ball is black.
The questions are: a) what is the probability of the blind to win
having the urn n balls b) what is the probability of the blind to win
when n tends to infinite. You can only use the four basic operations
(+-*/) and the number e. Then I find the probability of the blind to
lose using the relationship
P(A1.A2....An)=P(A1)*P(A2/A1)*....P(An/A1.A2...An-1).

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Clarification of Question by brandmath-ga on 23 Dec 2005 10:53 PST

(continuation)

The probability the blind to win will be P(A1.A2...An-1) where Ai is
the event of the i ball take off by the blind is not a black.
Logically if the n-1 balls are not black the last will be black. If
you develop that relationship wit n even and odd you will find those
two sequences.
Well this is the entire question, now could you help me?

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Clarification of Question by brandmath-ga on 26 Dec 2005 05:45 PST

I know the challenge is big  but do not give up. Let's solve the
problem before the end of 2005. Who will be the hero?

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Clarification of Question by brandmath-ga on 27 Dec 2005 04:59 PST

n greater our equal to 3

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Clarification of Question by brandmath-ga on 29 Dec 2005 07:33 PST

I thank mathtalk-ga for your intervention. Neverthless you also didn't
find an answer in terms of e. Would there be another solution where
the answer could be put in terms of e, easier then our solution?

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Clarification of Question by brandmath-ga on 31 Dec 2005 02:13 PST

I realized the misunderstanding. The true formula for Bn, for example,
is Bn=(2/3)8(4/5).....(n-1)/n and not your formula. The probability of
success for n=5 is really (2/3)*(4/5)and not that found with your
formula. Then lim n-> 00 SQRT(n)*Bn is 1,253189262 and not 0,8862.....
I think that we can put this limit in terms of e .

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Clarification of Question by brandmath-ga on 01 Jan 2006 08:49 PST

Dear mathtalk-ga. You are helping me and I thank you very much.
Neverthless, I have to show you what you did not undestand, may be
because my poor english. Please think about my following explanation.
When you have n odd balls in the urn the probability that the first
player win is NOT ONE. It is enough to see one case. Imagine n=3, if
the first player chose the black ball he will  lose do not matter what
occurs after. He only wins the game if he does not choose the black
ball before the last ball is taken off by himself from the urn.
I hope I have had vocabulary enough to make you understand. I would
like you continue help me because I realized you is very well prepared
in math.
Best regards brandmath

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Clarification of Question by brandmath-ga on 01 Jan 2006 18:22 PST

Dear mathtalk;
I can not change the problem. I think there must be an way of
transforming the product (for example (2/3)*(4/5)) in sum and to find
an aproximation for greater n in terms of e. If it is difficult this
is another subject. Remember that e= 1+1+ (1/2!)+(1/3!). I am sure of
one thing, there is a solution in terms of e although I couldn't find
itd yet.

One thing to notice is that the product of the limits of products of
both your sequences is zero.  That is,

(1/2)*(3/4)*(5/6)*...*[(2n-1)/(2n)] *(2/3)*(4/5)*(6/7)*...*[(2n)/(2n+1)] 

is equal to

(1/2)*(2/3)*(3/4)*(4/5)*(5/6)*(6/7)*...*[(2n-1)/(2n)]*[(2n)/(2n+1)]

which collapses to 1/(2n+1),

which clearly approaches zero as n approaches infinity.  Since the
product of the limits is zero, at least one of the individual limits
must be zero.  In reality, they both are.

Emprically, the product of the first sequence quickly converges to
0.564190/sqrt(n), and the product of the second sequence to
0.886227/sqrt(n), where n is the number of terms in the sequence.  I
don't know how to derive these expressions analytically, or whether
the approximate values 0.886227 and 0.564190 correspond to exact
numbers that can be written in terms of integers, pi, e, special
functions, etc.

Let A_n = (1/2)*(3/4)*(5/6)*...*(2n-1)/(2n)
and B_n = (2/3)*(4/5)*(6/7)*...*(2n)/(2n+1).

As rracecarr-ga points out, A_n * B_n simplifies to 1/(2n+1).

Also:  (1/2) < (2/3) < (3/4) < (4/5) < ...

Therefore A_n < B_n and B_n < 2 * A_{n+1}.

Since {A_n} and {B_n} both are strictly decreasing sequences of
positive numbers, they are convergent.  As rracecarr-ga pointed out,
the limit of their product is zero, so at least one of:

  LIMIT  A_n ,  LIMIT  B_n
  n->oo         n->oo

must be zero.  But the inequalities above establish:

  LIMIT  A_n  =  0  if and only if  LIMIT  B_n  =  0
  n->oo                             n->oo

Therefore both limits are zero.

As to the asymptotic behavior of SQRT(n)*A_n and SQRT(n)*B_n noticed
by rracecarr-ga, consider first the ratio:

  A_n     (1/2)*(3/4)*(5/6)*...*(2n-1)/(2n)
 ----- = -----------------------------------
  B_n     (2/3)*(4/5)*(6/7)*...*(2n)/(2n+1)

          (1*3)*(3*5)*(5*7)*...*((2n-1)*(2n+1))
       = ---------------------------------------
          (2*2)*(4*4)*(6*6)*...*( (2n) * (2n) )

whose limit is 2/pi.  See for example:

[Wallis product -- Wikipedia]
http://en.wikipedia.org/wiki/Wallis_product

Thus {SQRT(n)*A_n} and {SQRT(n)*B_n} are positive sequences:

 -  whose product converges to 1/2, and

 -  whose ratio converges to 2/pi.

From these facts we can deduce that:

  LIMIT SQRT(n)*A_n = SQRT( (1/2) * (2/pi) )
  n->oo
                    = 1/SQRT(pi)

                    = 0.56418958354775628694807945156077...

  LIMIT SQRT(n)*B_n = SQRT( (1/2) / (2/pi) )
  n->oo
                    = SQRT(pi)/2 

                    = 0.88622692545275801364908374167057...

A more careful analysis leads to this expansion and an explicit
expression in terms of the gamma function, also known as "Wallis'
Formula":

             1               
  A_n = ---------- * [ 1 - 1/(8n) + 1/(128n^2) - ... ]
        SQRT(pi*n)

           gamma(n + 1/2)
      = ---------------------
        SQRT(pi)*gamma(n + 1)

and similarly for B_n.


regards, mathtalk-ga

Not sure where you are going with the Clarification, brandmath-ga.

The original Question was what is the limit of A_n, resp. B_n.

The answer to that, as rracecarr-ga pointed out and I elaborated, is
zero for both sequences.

Now zero is equally well expressed in terms of e or pi, but really
it's much simpler than either.

If you wanted an exact expression for A_n, such as I gave in terms of
the gamma function, I don't think it can be much simpler than this. 
However there is an integral formula.

It might help to give more of an idea what the context (your
probability problem) is and how useful one or another formulation
might be to simplifying or solving the problem.

regards, mathtalk-ga

With respect to the Clarification of the original Question's context,
note that the probability of the first player winning is 1 (certainty)
if there are an odd number of balls in the urn, including the black
one.

If the number of balls in the urn is even, there is a probability
(tending to zero as the number of balls grows without limit) that the
first player may fail to win, by leaving the black ball in the urn
until last, in which case the second player cannot take a turn because
the second player is only allowed to draw a white ball.

Let p_n be the probability that this happens (first player fails to
win) with 2n balls in the urn.  Since this requires the first draw to
be white (and which will automatically be followed by the second draw
of white), we spot the following recurrence relationship because after
two white balls are gone, the resulting endgame has probability
p_{n-1} of the first player not winning:

  p_n = ((2n-1)/(2n)) * p_{n-1}

Since p_2 = 1/2, mathematical induction proves that:

  p_n = (1/2)*(3/4)* ... * ((2n-1)/(2n))

which is identical to the expression I previously called A_n without
knowing the context.  As rracecarr-ga and I have both concluded, this
expression tends to zero as n goes to infinity.

To summarize: the probability that the first player will win is
exactly 1 if there are an odd number of balls in the urn, and exactly
1 - p_n if there are an even number 2n of balls in the urn before the
first draw.

In either case the chances of the first player winning tend to 1 as
the number of balls in the urn increases without limit.

Bearing in mind that the exact form of p_n = A_n was given in terms of
the gamma function, and that the approximate behavior of this value is
1/SQRT(pi * n), I'd say that the issue of the probability of the first
player winning has been dealt with.

It might be interesting to pose a slightly different problem.  How
does the expected number of turns (draws) change as the number of
balls in the urn grows?  Note that the number of draws is uncertain
for an odd number of balls, despite the certainty that the first
player will eventually remove it.


regards, mathtalk-ga

The specific values requested were:

(a) the probability of the [first player] to win having [in] the urn n balls

(b) the probability of the [first player] to win when n tends to [infinity]

These values have been shown to be rational numbers.  Taking into
consideration the further request by brandmath-ga:

"You can only use the four basic operations (+-*/) and the number e."

it can be shown that the only such formulas which express rational
numbers use the number e in a trivial way.  For example, we can
express the value 1, which is the answer to (b), as e/e.

Specifically, the given operations and the number e will produce only
a rational function of the transcendental number e, that is an
expression p(e)/q(e) where p(x) and q(x) are polynomials.

Such an expression is equal to a rational number only if the
polynomials p(x) and q(x) are of equal degree and have corresponding
coefficients whose ratio is exactly that rational number.

In other words, p(e)/q(e) = r is a rational number if and only if p(x)
= r*q(x) as polynomials.

regards, mathtalk-ga

Let me try to clear up the confusion.  First I asserted that when an
odd number of balls are in the urn, one black ball together with an
even number of white balls, then the first player always wins.  I did
not assert that B_n as I formulated it was related to the probability
of winning in the cases where the urn contains an odd number of balls,
although I did related A_n as I formulated it to the probability of
winning where the urn contains an even number of balls.

In support of this interpretation of the rules for an odd number of
balls, let me point out brandmath-ga's statement that "Whatever
happens, the last ball must be taked off by the blind."  brandmath-ga
refers to the first player as "blind" in respect of taking a ball from
the urn without knowing in advance its color. Since the second player
will always be able to remove one white ball for every white ball
taken by the first player when there an even number of white balls, in
this circumstance play terminates only when the black ball is finally
taken by the first player.

Second, brandmath-ga raises the possibility that a slightly different
"formula" for B_n may lead to a different limit:

  (2/3) * (4/5) * ... * (n-1)/n

I assume the values allowed for n in this formula are restricted to
odd integers greater than 1, since otherwise the "pattern" suggested
would not make sense.  That is, judging by the first two factors, all
the numerators in this product are even and all the denominators are
odd (including the final denominator n).

Is this really a different formula?  Actually if odd n = 2k+1, then it
agrees with what I previously labelled B_k.

brandmath-ga claims that the limit of the new formula time SQRT(n)
differs from that previously given and holds out the hope that somehow
the new limit may be said to involve the transcendental number e.

"Then lim n-> 00 SQRT(n)*Bn is 1,253189262 and not 0,8862.....
I think that we can put this limit in terms of e ."

However the observed change in limit is merely the factor of SQRT(2)
introduced by changing the indexing from k to 2k+1, that is using
SQRT(n) = SQRT(2k+1) rather than SQRT(k):

SQRT(2) * (SQRT(PI)/2) = 1.414... * 0.8862...

                       = 1.2533141373155002512078826424055...


For my analysis the sequence B_n was useful in determining the limits
of both {SQRT(n)*A_n} and {SQRT(n)*B_n} together.  Changing the
formulation to involve {SQRT(2k+1)*A_k} and {SQRT(2k+1)*B_k} would not
significantly alter the results, other than to increase both limits by
a factor of SQRT(2).


regards, mathtalk-ga

Yes, I obviously misunderstood when I thought that the "game" ended
when the black ball is drawn by the first player or when the second
player is unable to have a turn because the only remaining ball is
black.  I understood your statement:

"Whatever happens, the last ball must be taked off by the blind. The
blind will win if the last ball is black."

in the sense of the "last ball" drawn by the first player, not the
last ball in the urn.

Although your new description of the rules of the "game" makes sense
for an odd number of balls, and allows for the first player to either
win or not win, it is now unclear what the rules are with an even
number of balls.  If the players take turns drawing balls, then an
even number of balls implies that the last ball will generally be
taken off by the second player.

I would suggest the following "rules" (though of course you are the
expert on what problem you are trying to solve).  An urn has n balls,
one of which is black and the rest white.  Two players take turns
drawing one ball at a time.  The first player draws "randomly" without
looking at the color of the ball.  The second player draws a white
ball if one is available.  The game ends when either player is unable
to take their turn, either because the first player finds no ball is
left in the urn or the second player finds no white ball is left in
the urn.

You seem to define the "winning" probability as the chance that the
first player's last draw will be a black ball.  To illustrate, we
consider several values of n, both even and odd.

If n = 2, Pr("win") = 1/2.
If n = 3, Pr("win") = 2/3.
If n = 4, Pr("win") = (3/4)*(1/2).
If n = 5, Pr("win") = (4/5)*(2/3).

More generally when n > 3, the chance p_n of a "win" (first player's
last draw is the black ball) is (n-1)/n times p_{n-2}, because to win
the first player's first draw must be a white ball, and after the
second player draws another white ball, play is reduced to the game
with an urn containing n-2 balls.

Now we have for the cases of an even number n=2k of balls and an odd
number n=2k+1 of balls:

  p_{2k} = A_k

 p_{2k+1} = B_k

as I defined those sequences previously, using a different subscript:

    A_k = (1/2)*(3/4)*(5/6)*...*(2k-1)/(2k)
and B_k = (2/3)*(4/5)*(6/7)*...*(2k)/(2k+1).

I have nothing further to add about the asymptotic behavior of these sequences.

If your goal to have an Answer that involves the number e in some
particularly simple yet useful way, you will need to change the
Question.  For example, one often cited puzzle is the chance that if n
people have hats that are randomly redistributed (permuted), none will
get their own hat.  As n tends to infinity, this probability tends to
1/e.

regards, mathtalk-ga

The values of e and pi, both transcendental numbers, are related such
as by Euler's famous formula:

e^(i * pi) = -1

but as far as is known there is no simple algebraic relation between them.

If you absolutely need to drag the constant e into the picture, then I
would suggest the most natural approach may be using this definition
of the gamma function:

gamma(N) = INTEGRAL e^(-x) * x^(N-1) dx FROM x = 0 to +oo

[gamma function -- NIST]
http://www.nist.gov/dads/HTML/gammaFunction.html

in combination with the exact values for A_n given earlier:

           gamma(n + 1/2)
  A_n = ---------------------
        SQRT(pi)*gamma(n + 1)

Note that I have reverted to the definition of A_n indexed by n, so
that the final factor of A_n is (2n-1)/(2n).

In this context A_n times B_n is 1/(2n+1), as rracecarr-ga pointed
out.  You should be able to work out the analogous formula for B_n in
terms of the gamma function using B_n = 1/(A_n * (2n+1)).

Incidentally you can check these formulas by using the Windows
"calculator" applet or similar software, because it will evaluate x!
for non-integer values of x consistent with the identity gamma(x+1) =
x!.


best wishes, mathtalk-ga