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

Here's one rough approach which works for $n$ sufficiently large. The idea is that for large $n$, we have that
$$
F_n \approx \phi^n/\sqrt{5}
$$
since the conjugate root is less than one, and so it tends to zero.
So consider now the generating function
$$
G(q) = \sum_{n=0}^\infty \frac{q^n}{F_n}
$$
We note that your sum approaches $\big(q\frac{d}{dq}G(q)\big)_{q=1}$ and so we just need to evaluate that expression. Using the approximating above, we see that
$$
G(q) \approx \sum_{n=0}^\infty \sqrt{5}\frac{q^n}{\phi^n} = \sqrt{5}\frac{\phi}{\phi - q}
$$
and so taking the derivative we get
$$
q\frac{d}{dq}G(q) \approx \frac{\sqrt{5}\phi}{(\phi - 1)^2}
$$
which, when evaluated at $q = 1$ yields $9.4721\ldots < 13$
There are obviously some details that would probably need to be filled in here (how good are the approximations?), but this could give you a rough start.
Edit
Here is a full proof. We first define $G_n(q) = \sum_{k=0}^n \frac{nq^n}{F_n}$ and $G(q) = \sum_{k=0}^\infty \frac{nq^n}{F_n}$. Clearly, $G_n(q) < G(q)$ (if $q > 0$). So our goal is to bound $G(1)$. Define $D = q\frac{d}{dq}$. We have:
$$
\begin{align}
G(q) &= D\sum_{k=0}^\infty \frac{q^n}{F_n} \\
&= D\sum_{k=0}^\infty \frac{\sqrt{5}q^n}{\phi^n - \phi'^n} \\
&= D\sum_{k=0}^\infty \frac{\sqrt{5}q^n}{\phi^n\big(1 - (\phi'/\phi)^n\big)}
\end{align}
$$
The point now is that we always have that
$$
\frac{1}{1 - (\phi'/\phi)^n} < \frac{1}{1 - (\phi'/\phi)^2}
$$
since this is equivalent to $(\phi'/\phi)^n < (\phi'/\phi)^2$ which is vacuously true if $n$ is odd (the LHS is negative) and true if $n$ is even since $|\phi'/\phi| < 1$. Thus
$$
G(q) = D\sum_{k=0}^\infty \frac{\sqrt{5}q^n}{\phi^n\big(1 - (\phi'/\phi)^n\big)} < \frac{\sqrt{5}}{1 - (\phi'/\phi)^2}D\sum_{k=0}^\infty \frac{q^n}{\phi^n}
$$
from which we have (looking at the previous argument) that
$$
G_n(1) < G(1) < \frac{\sqrt{5}}{1 - (\phi'/\phi)^2}\frac{\phi}{(\phi - 1)^2} \approx 11.08 < 13
$$
