[Guest]

Prove which of the following have a limit of zero and which do not. Where a non-zero limit exist what is it?

1. (1 + 1/2)(1 - 1/2) * (1 + 1/3)(1 - 1/3) * (1 + 1/4)(1 - 1/4)*.....

2. cos(1) * cos(1/2) * cos(1/3) * cos(1/4) * .....

[Guest]

for 1) we will eventually get to (1 + 1/infinite)(1 -1/infinite) which equals (1 +0)(1-0) = 1 so i'd say it's limit is one

for 2) we will eventually get to cos(1/infinite) which equals cos(0) which equals 1 so I'd say it's limit is one

for both cases it never gets over 1 just closer and closer to it

ps: how do you type the infinity symbol?

[unreality]

Prove which of the following have a limit of zero and which do not. Where a non-zero limit exist what is it?

1. (1 + 1/2)(1 - 1/2) * (1 + 1/3)(1 - 1/3) * (1 + 1/4)(1 - 1/4)*.....

2. cos(1) * cos(1/2) * cos(1/3) * cos(1/4) * .....

using computer programs to approximate the limits:

starting with #2

from cos(1) all the way to cos(1/100,000,000), the product is .3885361544 or so and by the time you get with numbers so high, the 1/part makes it so close to 0 that the cosine comes out to be basically 1. So it's not really changing after this point. I don't think the number is special (although it's close to pi eighths), it's just based on the very first number cos(1) which is about .54 and gets a little smaller but each consecutive cosine gets closer and closer to 1 so it's like starting with about .54 and multiplying by numbers that get bigger and bigger, approaching 1 from below but not quite, so it gets a little smaller (dropping a bit below .39) but as a limit it'll eventually be multiplying by essentially 1's and not changing. Sure with each multiplication, the factor is less than 1, therefore the number does shrink a little bit with each mult., but the amount it shrinks by gets smaller and smaller.

I don't know how to prove limit stuff when you have infinite products (sums are easier) but I'm gonna say with confidence that #2 goes to a limit.

As for #1, the limit is 1/2. Yes first I wrote a computer program to figure it out haha, but it does make sense:

(1 + 1/n)(1 - 1/n) = (1 + 1/n - 1/n - 1/n^2) = 1 - 1/n^2

so it's the infinite product of ( 1 - 1/nn) as n goes from 2 to infinity

(1 - 1/4) (1 - 1/9) (1 - 1/16) (1 - 1/25) etc

Multiply the first two together

(1 - 1/4 - 1/9 + 1/36) = (1 - 9/36 - 4/36 + 1/36) = (1 - 12/36) = 1 - 1/3 = 2/3

2/3 by (1 - 1/16) = 2/3 - 1/24

it's 1 - 1/3 - 1/24

now take that (which is 1 - 9/24 = 1 - 3/8 = 5/8) and multiply by (1-1/25)

1 - 1/3 - 1/24 - 5/(8*25)

equals

1 - 1/3 - 1/24 - 1/40

And etc.

Now it's easy to prove that 1 - 1/4 - 1/8 - 1/16 - 1/32 - 1/64 - 1/128 - ... etc will go to 1/2

The obvious conclusion is that 1/4 + 1/8 + 1/16 etc isn't the only infinite series to add to 1/2

[Guest]

Converges on 1/2

Converges on 0.388556 but I can't solve with a proof

[Guest]

the formula for the nth partial product is (n+1)/(2n)

the limit of this as n approaches infinity is 1/2 because the n+1 becomes less and less significant.

unfortunately, this one has no simple answer. however, here it is to a ridiculous number of digits...

0.3885361533351758591843295765490569412676615819170160388191651983467795065937849429335651032060616362

[Guest]

is 1/2.

It's equal to (1X3/2^2) X (2X4/3^2) X (3X5/4^2） X （4X6/5^2)X...[(n-1)X(n+1)/n^2]

Everything else cancels out when n is sufficeiently large except one 2 in the denominator.

[Guest]

For 1*(1-1/2)*(1+1/2)*(1-1/3)*(1+1/3)... The answer is 0.500104

For 2*cos(1)*cos(1/2)*cos(1/3)*cos(1/4)... The answer is 0.777164

Edited June 17, 2010 by parik

[Guest]

1) series will limit to 1/6.

Series can be written as [1-1/k^2][1-1/(k+1)^2] which simplifies to (k-1/k)(k+2/k+1) when we put values the whole series is cancelled with remaining 1/6.

2) the series can be separated by 2cosAcosB and it will zero as 2^n comes in dinominator.