[Guest]

Diverge or converge?

(1 + 1/2) * (1 + 1/4) * (1 + 1/8) * (1 + 1/16) * ...

(A proof would be nice )

[Guest]

Not even sure if I can do this but:

(1+1/2)*(1+1/4)*(1+1/8)*(1+1/16) * ... = (3/2)(5/4)(9/8)(17/16)... = (3*5*9*17....)/(2*4*8*16....)

The top increases faster than the bottom so it diverges.

[Guest]

this equation can be first described as:

(1+ (1/2^t)) * (1+ (1/2^t+1)) * (1+ (1/2^t+2)) * (1+ (1/2^t+3))....

Well, lets look at what happens to the equation as t approaches infinity.

The value of 1/(2^t) becomes close to zero because the denominator becomes so large. Thus, the "last" (theoretically) term is just 1. So after that, we know that the sum will never increase any more. Therefore, the sum has a finite value, therefore the equation converges. Q.E.D.

[unreality]

Diverge or converge?

(1 + 1/2) * (1 + 1/4) * (1 + 1/8) * (1 + 1/16) * ...

(A proof would be nice )

using only the knowledge that d/dx[e^x] = e^x because of its exponential nature, and choosing e as the base to make it exactly equal (other bases would have a constant multiplier), without yet knowing what the value of e actually is.

Now we know that ln x is the logarithm base e of x, the inverse operation. Now we want to know d/dx[lnx]

y = ln x

e^y = x

d/dx[e^y] = x

e^y * dy/dx = 1 by the chain rule, so dy/dx = 1/e^y

but y = ln x so dy/dx = 1/e^(lnx)

and e^(lnx) = x

ergo

dy/dx = 1/x

where dy/dx is the derivative of the natural logarithm.

so d/dx[lnx] = 1/x

But from the definition of the derivative.

d/dx[lnx] = lim(h->0) of ln(x+h) - ln(x) all over h

= lim(h->0) of ln((x+h)/x) / h = lim(h->0) of ln(1 + h/x) / h

alnb = ln(b^a) so:

d[lnx]/dx = lim(h->0) of ln(1+h/x)^(1/h)

replace with h with 1/n

dy/dx = lim(n->infinity) of ln(1 + 1/nx)^n

but we already know that dy/dx = 1/x

so now pull the natural log out of the limit cuz it's continuous, and raise e to both sides

e^(1/x) = lim(n->infinity) of (1+1/nx)^n

to find e, plug in 1 for x

e^(1/1) = e = lim( n -> inf) of (1 + 1/n)^n

but what does it have to do with the infinite product of (1+1/2^x) ?

if we consider the ^n as a multiplication of something by itself 'n' times, and find limit as n -> inf, it's like this:

(1+1/1)

(1+1/2) * (1+1/2)

(1+1/3) * (1+1/3) * (1+1/3)

etc as n approaches infinity.

the difference is that our product can't be raised to a power because it's not the same thing being multiplied by itself. But we can find an average value that can reduce the infinite product into a nice exponential expression? Let's start with n=1, n=2, etc, and see if we can define a limit as n approaches infinity. For the first case, of (1+1/2) * … * (1+2^(-n)) I mean, we get just the one term:

(1+1/2)

for n=2, we get

(1+1/2)(1+1/4) = 1 + 1/2 + 1/4 + 1/8

for n=3, multiply that by (1+1/8) and distribute, multiply all by the one first, then again by the 1/8, so we've got

1 + 1/2 + 1/4 + 1/8 + (1/8 + 1/16 + 1/32 + 1/64)

the two eights combine to make a fourth

1 + 1/2 + 1/4 + 1/4 + 1/16 + 1/32 + 1/64

which combines with the other fourth to make a half

1 + 1/2 + 1/2 + 1/16 + 1/32 + 1/64

which combine for a 1

1 + 1 + 1/16 + 1/32 + 1/64

2 + 1/16 + 1/32 + 1/64

It all won't be that pretty. When n=4, we're multiplying by (1+1/16), but with n=1,2,3 we should have some sufficient data to detect a pattern.

now for n=1, we can represent it exponentially like this:

(1+1/2)^1

but how to represent n=2 as a single term raised to a power?

That was 1 + 1/2 + 1/4 + 1/8, which is 8 + 4 + 2 + 1 all over 8, or 15/8, nearly 2.

If we were to represent that as a square, which is what the pattern would suggest we do, it would have to be

(sqrt15 / sqrt8) ^ 2

whereas the first term was (3/2)^1 or (sqrt9 / sqrt4) ^ 1

For this to continue working, then this:

2 + 1/16 + 1/32 + 1/64

has to be a perfect cube.

it evaluates to

128 + 4 + 2 + 1, all over 64

135 / 64

64 is a perfect cube… but 135 isnt

Dang it, I guess I'm at a loss right now, and I thought all that e/ln stuff would be useful. Maybe it will be for someone if they want to have a crack at it. I wrote a little java program for it…

converge on the value 2.3842310290313713

I'll think more on this…

[Guest]

Let P(n):=(1+1/2)*(1+1/4)*...*(1+(1/2)^n), L(x):=log x, and S(n):=L(1+1/2)+L(1+1/4)+...+L(1+(1/2)^n). Then L(P(n))=S(n). Since L(x) is continuous for x>0, it suffices to show that S(n) converges.

Observe that L(1+(1/2)^n)=L(2^n+1)-L(2^n).

Also, observe that L'(x)=1/x for x>0.

So by the Mean Value Theorem, L(1+(1/2)^n)<(1/2)^n for each n>0.

In particular S(n)<1/2+1/4+...+(1/2)^n.

Since 1/2+1/4+...+(1/2)^n+... is geometric series, we get that S(n) converges.

In fact, P(infinity)=E(S(infinity)), where E(x):=exp(x) denotes the exponential function base "e."

Edited January 17, 2010 by five0to180

[Guest]

nice answer!

[Guest]

Diverge or converge? (1 + 1/2) * (1 + 1/4) * (1 + 1/8) * (1 + 1/16) * ... (A proof would be nice )

I think it converges.

Your infinite product converges if and only if the series 1/2 + 1/4 + 1/8 + ... converges*,

which it does because it cannot exceed 1.

(Think of a one-meter segment. If you step ½ meter, then ¼ meter, etc. each step is half of the remaining distance and so you can never travel more than one meter.)

* In general, an inifinte product [1 + f(n)] converges iff the infite sum of f(n) converges (can supply proof, but didn't include).

[Guest]

Diverge or converge? (1 + 1/2) * (1 + 1/4) * (1 + 1/8) * (1 + 1/16) * ... (A proof would be nice )

I think it converges.

Your infinite product converges if and only if the series 1/2 + 1/4 + 1/8 + ... converges*,

which it does because it cannot exceed 1.

(Think of a one-meter segment. If you step ½ meter, then ¼ meter, etc. each step is half of the remaining distance and so you can never travel more than one meter.)

* In general, an inifinte product [1 + f(n)] converges iff the infite sum of f(n) converges (can supply proof, but didn't include).

Your final remark is false: Consider the sequence f(n)=1/n. We know that the infinite sum of 1/n diverges. On the other hand, if we define P(n):=(1+1/1)*(1+1/2)*...*(1+1/n). Then P(n) converges using the same argument I made above. The reason why "infinite product [1+f(n)]" can converge while "infinite sum f(n)" may not is because the log function "slows down" the sequence [1+f(n)]. If you look at the growth of log(x), one sees that it goes to infinity as x gets large, but quite slowly since it's derivative is 1/x. In specific, the key ingredient in the proof I gave above was that the derivative of log(x) was not only bounded for x>1, but decreasing as x gets large.

Edited January 18, 2010 by five0to180

[Guest]

Your final remark is false: Consider the sequence f(n)=1/n. We know that the infinite sum of 1/n diverges. On the other hand, if we define P(n):=(1+1/1)*(1+1/2)*...*(1+1/n). Then P(n) converges using the same argument I made above. The reason why "infinite product [1+f(n)]" can converge while "infinite sum f(n)" may not is because the log function "slows down" the sequence [1+f(n)]. If you look at the growth of log(x), one sees that it goes to infinity as x gets large, but quite slowly since it's derivative is 1/x. In specific, the key ingredient in the proof I gave above was that the derivative of log(x) was not only bounded for x>1, but decreasing as x gets large.

actually, xamdam is absolutely correct. the infinite product of (1+1/n) after "x" terms is equal to x+1. Thus, it diverges.

[bonanova]

2.384231029.....

Proof: calculation

[Guest]

Your final remark is false: Consider the sequence f(n)=1/n. We know that the infinite sum of 1/n diverges. On the other hand, if we define P(n):=(1+1/1)*(1+1/2)*...*(1+1/n). Then P(n) converges using the same argument I made above. The reason why "infinite product [1+f(n)]" can converge while "infinite sum f(n)" may not is because the log function "slows down" the sequence [1+f(n)]. If you look at the growth of log(x), one sees that it goes to infinity as x gets large, but quite slowly since it's derivative is 1/x. In specific, the key ingredient in the proof I gave above was that the derivative of log(x) was not only bounded for x>1, but decreasing as x gets large.

Your P(n)= (1 + 1/1)*(1+1/2)*(1 + 1/3)*(1 + 1/4)...*(1+1/n)

= (2) * (3/2) * (4/3) * (5/4) * (6/5) * ...

these partial products are 2,3,4,5,6,…

Aren't these diverging?