show that $\frac{1}{F_{1}}+\frac{2}{F_{2}}+\cdots+\frac{n}{F_{n}}<13$ Let $F_{n}$ is Fibonacci number,ie.($F_{n}=F_{n-1}+F_{n-2},F_{1}=F_{2}=1$)
show that
$$\dfrac{1}{F_{1}}+\dfrac{2}{F_{2}}+\cdots+\dfrac{n}{F_{n}}<13$$
if we use 
Closed-form expression
$$F_{n}=\dfrac{1}{\sqrt{5}}\left(\left(\dfrac{1+\sqrt{5}}{2}\right)^n-\left(\dfrac{1-\sqrt{5}}{2}\right)^n\right)$$
$$\dfrac{n}{F_{n}}=\dfrac{\sqrt{5}n}{\left(\left(\dfrac{1+\sqrt{5}}{2}\right)^n-\left(\dfrac{1-\sqrt{5}}{2}\right)^n\right)}$$
Well and now I'm stuck and don't know how to proceed

Let $\phi=\frac{1+\sqrt{5}}{2}$. Then $\phi^2=\phi+1$.
By induction, we have $F_n > \phi^{n-2}$ for all $n\ge 1$, and so
$$
\sum_{n=1}^{N} \frac{n}{F_n}
<
\sum_{n=1}^{\infty} \frac{n}{F_n}
< \sum_{n=1}^{\infty} \frac{n}{\phi^{n-2}}
= \sum_{n=1}^{\infty} \frac{n\phi^2}{\phi^{n}}
= \phi^2 \sum_{n=0}^{\infty} \frac{n}{\phi^{n}}
= \frac{\phi^3}{(1-\phi)^2}
= \frac{1+2\phi}{2-\phi}
< 13
$$
because $\phi< \dfrac53$.
I've used that $\phi^2=\phi+1$ and so $\phi^3=\phi\phi^2=\phi^2+\phi=2\phi+1$.
(The value of the last fraction above is $\approx 11.09$. It is the same as Simon's bound.)
