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

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

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

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

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}

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?