This is a typical proof one may see in a undergraduate introduction to proofs mathematics course.

Proposition 1 Let the constant {R} be defined such that

\displaystyle R:=\frac{1}{F_{1}}+\frac{1}{F_{2}\cdot F_{1}}+\frac{1}{F_{3}\cdot F_{2}\cdot F_{1}}+\frac{1}{F_{4}\cdot F_{3}\cdot F_{2}\cdot F_{1}}\cdots

and {F_{n}} are the Fibonacci numbers where {F_{n+k}>F_{k}} for {n>0}. Then {R} is irrational.

Proof: (Contradiction) Let the constant {R} be defined as

\displaystyle R:=\frac{1}{F_{1}}+\frac{1}{F_{2}\cdot F_{1}}+\frac{1}{F_{3}\cdot F_{2}\cdot F_{1}}+\frac{1}{F_{4}\cdot F_{3}\cdot F_{2}\cdot F_{1}}\cdots.

Note {F_{n}} are the Fibonacci numbers where {F_{1,2}=1} and {F_{n+k}>F_{k}} where {k,n\in\mathbb{N}}.

Assume {R} is rational; {R\in\mathbb{Q}}. This implies

\displaystyle R=\frac{p}{q},

{p,q\in\mathbb{Z}} and {q\neq0} are fixed integers. Next, lets consider the difference for {n\geq1}

\displaystyle R-(\frac{1}{F_{1}}+\frac{1}{F_{2}\cdot F_{1}}+\frac{1}{F_{3}\cdot F_{2}\cdot F_{1}}+\cdots+\frac{1}{F_{n}\cdot F_{3}\cdot F_{2}\cdot F_{1}}).

Note, the difference is greater than zero:

\displaystyle 0<R-(\frac{1}{F_{1}}+\frac{1}{F_{2}\cdot F_{1}}+\frac{1}{F_{3}\cdot F_{2}\cdot F_{1}}+\cdots+\frac{1}{F_{n}\cdot F_{3}\cdot F_{2}\cdot F_{1}}).

Now, we can estimate this difference. From the definition of {R}, we can see that

\displaystyle R-(\frac{1}{F_{1}}+\frac{1}{F_{2}\cdot F_{1}}+\frac{1}{F_{3}\cdot F_{2}\cdot F_{1}}+\cdots+\frac{1}{F_{n}\cdot F_{3}\cdot F_{2}\cdot F_{1}})

\displaystyle =\frac{1}{F_{n+1}\cdot F_{n}\cdot F_{3}\cdot F_{2}\cdot F_{1}}+\frac{1}{F_{n+2}\cdot F_{n+1}\cdot F_{n}\cdot F_{3}\cdot F_{2}\cdot F_{1}}+\cdots

\displaystyle =\frac{1}{F_{n+1}\cdot F_{n}\cdot F_{3}\cdot F_{2}\cdot F_{1}}(1+\frac{1}{F_{n+2}}+\frac{1}{F_{n+3}\cdot F_{n+2}}+\cdots)

\displaystyle <\frac{1}{F_{n+1}\cdot F_{n}\cdot F_{3}\cdot F_{2}\cdot F_{1}}(1+\frac{1}{F_{2}}+\frac{1}{F_{3}\cdot F_{2}}+\cdots)

\displaystyle =\frac{1}{F_{n+1}\cdot F_{n}\cdot F_{3}\cdot F_{2}\cdot F_{1}}(\frac{1}{F_{1}}+\frac{1}{F_{2}\cdot F_{1}}+\frac{1}{F_{3}\cdot F_{2}\cdot F_{1}}+\cdots)

\displaystyle =\frac{R}{F_{n+1}\cdot F_{n}\cdot F_{3}\cdot F_{2}\cdot F_{1}}

So, we have arrived at the following bounds

\displaystyle 0<R-(\frac{1}{F_{1}}+\frac{1}{F_{2}\cdot F_{1}}+\frac{1}{F_{3}\cdot F_{2}\cdot F_{1}}+\cdots+\frac{1}{F_{n}\cdot F_{3}\cdot F_{2}\cdot F_{1}})

\displaystyle <\frac{R}{F_{n+1}\cdot F_{n}\cdot F_{3}\cdot F_{2}\cdot F_{1}}.

Since {R=\frac{p}{q}},

\displaystyle 0<\frac{p}{q}-(\frac{1}{F_{1}}+\frac{1}{F_{2}\cdot F_{1}}+\frac{1}{F_{3}\cdot F_{2}\cdot F_{1}}+\cdots+\frac{1}{F_{n}\cdot F_{3}\cdot F_{2}\cdot F_{1}})

\displaystyle <\frac{R}{F_{n+1}\cdot F_{n}\cdot F_{3}\cdot F_{2}\cdot F_{1}}.

Let {\frac{1}{F_{1}}+\frac{1}{F_{2}\cdot F_{1}}+\frac{1}{F_{3}\cdot F_{2}\cdot F_{1}}+\cdots+\frac{1}{F_{n}\cdot F_{3}\cdot F_{2}\cdot F_{1}}=\frac{A}{F_{n}\cdot F_{3}\cdot F_{2}\cdot F_{1}},A\in\mathbb{Z}}. We obtain

\displaystyle 0<\frac{p}{q}-\frac{A}{F_{n}\cdot F_{3}\cdot F_{2}\cdot F_{1}}<\frac{R}{F_{n+1}\cdot F_{n}\cdot F_{3}\cdot F_{2}\cdot F_{1}}.

Multiplying the inequality by {F_{n}\cdot F_{3}\cdot F_{2}\cdot F_{1}\cdot q} we obtain

\displaystyle 0<p\cdot F_{n}\cdot F_{3}\cdot F_{2}\cdot F_{1}-Aq<\frac{Rq}{F_{n+1}}.

Note that {p\cdot F_{n}\cdot F_{3}\cdot F_{2}\cdot F_{1}-Aq\in\mathbb{Z}^{+}} and so this further restricts the bounds as

\displaystyle 1\leq p\cdot F_{n}\cdot F_{3}\cdot F_{2}\cdot F_{1}-Aq<\frac{Rq}{F_{n+1}}

or

\displaystyle F_{n+1}<Rq.

Since we can pick {n} large enough such that

\displaystyle F_{n+1}>Rq.

This is a contradiction and thus our assumption must be wrong. {R} must be irrational; {R\notin\mathbb{Q}}. \Box

And that’s it! How would you have proven this?

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