Fermat Number Fractions

Calculus Level 3

Let $F_n$ denote the $n^\text{th}$ Fermat number, $F_n = 2^{2^n} + 1$ . Find $\displaystyle \large \prod_{n=0}^\infty \left( 1 - \dfrac1{F_n} \right) .$

The answer is 0.5.

1 solution

Michael Mendrin
Feb 13, 2016

This converges EXTREMELY fast. After just $7$ terms, it's $0.5$ followed by $37$ zeroes.

Edit: But we can see how this works inductively

$\left( \dfrac { { 2 }^{ { 2 }^{ n } } }{ 2\left( { 2 }^{ { 2 }^{ n } }-1 \right) } \right) \left( 1-\dfrac { 1 }{ { 2 }^{ { 2 }^{ n } }+1 } \right) =\left( \dfrac { { 2 }^{ { 2 }^{ n+1 } } }{ 2\left( { 2 }^{ { 2 }^{ n+1 } }-1 \right) } \right)$

so we can see that the limit value is just $\dfrac{1}{2}$

you're right,Fermat numbers get really big,real fast

Hamza A - 5 years, 4 months ago

See induction proof

Michael Mendrin - 5 years, 4 months ago

you're really good at doing solutions with a few lines :)

Hamza A - 5 years, 4 months ago

@Hamza A – Let me let on a secret. First, I use up pages and pages in working out a problem and then the solution, and then I make it as short as possible because it's so much trouble to write it out in LaTex. By the time I'm through, everybody thinks I must be a genius.

Michael Mendrin - 5 years, 4 months ago

@Michael Mendrin – Lol,you really are though :)

Hamza A - 5 years, 4 months ago

@Hamza A – It was Thomas Edison that said that genius is 1% inspiration and 99% perspiration. Or something like that.

Michael Mendrin - 5 years, 4 months ago

@Michael Mendrin – yeah,except most of his patents were stolen :(,he was good at business ,not inventing

Hamza A - 5 years, 4 months ago

Fermat Number Fractions

The answer is 0.5.

1 solution

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