$$\frac{1}{1}+\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\dots$$
The Fibonacci sequence is defined as
$$F_n=F_{n-1}+F_{n-2}\quad\text{(where $F_1=F_2=1$)}$$
Where the sum we want to find is \(S_\infty\)
$$S_n=\frac{1}{F_1}+\frac{1}{F_2}+\frac{1}{F_3}+\dots+\frac{1}{F_n}$$
It follows that for very large \(n\)
$$F_n:F_{n+1}=1:\varphi$$
So if we divide the series into two - the sum of the first finite \(n\) reciprocals and the rest.
$$ S_\infty=S_n+\left(\frac{1}{F_{n+1}}+\frac{1}{F_{n+2}}+\frac{1}{F_{n+3}}+\dots\right) $$
We know the value of \(S_n\) is finite, and the rest approximates to a geometric sequence with common ratio \(\frac{1}{\varphi}\), its sum can be evaluated to a finite value with the geometric series formula.
$$ \begin{align*} \frac{1}{F_{n+1}}+\frac{1}{F_{n+2}}+\frac{1}{F_{n+3}}+\dots&\approx\frac{1}{F_{n+1}}+\frac{1}{\varphi}\frac{1}{F_{n+1}}+\frac{1}{\varphi^2}\frac{1}{F_{n+1}}+\dots\\ &=\frac{1}{F_{n+1}}\frac{1}{1-1/\varphi} \end{align*} $$
Therefore, \(S_\infty\) must also be finite.
The \(n\)th term of the Fibonacci series can be calculated with
$$F_n=\frac{\varphi^n-(-1/\varphi)^n}{\sqrt{5}}$$