Can anyone come up with a proof for...

Precision

Can anyone come up with a proof for...

The sum from n=0 to infinity of the term, (2^n) / ( (2^(2^n)) + 1 ) converges to a value equal to 1

This looks like,

1/3 + 2/5 + 4/17 + 8/257 + ..... = 1

Mellor

I think I did this or something like it at LC.
Can remeber now at all, the general equation of the forumale is needed, I think

Michael Collins

Don't think you would have done this in the Leaving Cert, something similar alright, but this raises a power to a power, a bit too complicated for the LC I'd say.

As for solving it, I'm not sure where you'd start. Have you had any success OP?

seandoiler

Precision wrote:

Can anyone come up with a proof for...

The sum from n=0 to infinity of the term, (2^n) / ( (2^(2^n)) + 1 ) converges to a value equal to 1

This looks like,

1/3 + 2/5 + 4/17 + 8/257 + ..... = 1

well not sure how accurate this is but what i did was to consider the partial sums
S_0=1/3
S_1=11/15
S_2=247/255
etc

now of course the infinite sum is the limit of the partial sums, so we need only find an expression for the partial sums.....keep the partial sums in the form a/b and we can see that the denominator for the nth partial sum is 2^(2^(n+1))-1, while the numerator is 2^(2^(n+1))-1 - 2^(n+1) and so we have

S_n = [2^(2^(n+1))-1 - 2^(n+1)] / [2^(2^(n+1))-1] = 1 - [2^(n+1) / (2^(2^(n+1))-1)]

we now calculate the limit as n goes to infinity of the term S_n and since (2^(2^(n+1))-1) goes to infinity much faster than 2^(n+1), we can conclude that the limit in this case is 1

hence the sum of the infinite series is 1

Precision

Thanks!

seandoiler wrote:

well not sure how accurate this is but what i did was to consider the partial sums
S_0=1/3
S_1=11/15
S_2=247/255
etc

now of course the infinite sum is the limit of the partial sums, so we need only find an expression for the partial sums.....keep the partial sums in the form a/b and we can see that the denominator for the nth partial sum is 2^(2^(n+1))-1, while the numerator is 2^(2^(n+1))-1 - 2^(n+1)

I went through it as you said... considering the partial sums.
S_0=1/3
S_1=11/15
S_2=247/255
S_3=65519/65535
S_4=4294967263/4294967295
Putting the numerators into a series starting with the zeroth term,
1, 11, 247, 65519, 4294967263,... n'th term is, 2^(2^(n+1))-1-(2^(n+1))
Putting the denominators into a series starting with the zeroth term,
3, 15, 255, 65535, 4294967295,... n'th term is, 2^(2^(n+1))-1

seandoiler wrote:

and so we have

S_n = [2^(2^(n+1))-1 - 2^(n+1)] / [2^(2^(n+1))-1] = 1 - [2^(n+1) / (2^(2^(n+1))-1)]

we now calculate the limit as n goes to infinity of the term S_n and since (2^(2^(n+1))-1) goes to infinity much faster than 2^(n+1), we can conclude that the limit in this case is 1

hence the sum of the infinite series is 1

Excellent!

Michael Collins

And I take it you just had to construct the Sn for the numerator and denominator by inspection, yeh?

seandoiler

Michael Collins wrote:

And I take it you just had to construct the Sn for the numerator and denominator by inspection, yeh?

yes this is what i did for the denominator, the numerator is easy to spot as you just take an increasing power of two from the denominator value at each stage