TITLE: show that $\frac{1}{F_{1}}+\frac{2}{F_{2}}+\cdots+\frac{n}{F_{n}}<13$
QUESTION [14 upvotes]: 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

REPLY [6 votes]: Lemma
$$F_{n}\ge\dfrac{n(n+1)(n+2)}{42},n\ge 5$$
proof:use  induction, since
$$\dfrac{1}{42}[k(k+1)(k+2)+(k+1)(k+2)(k+3)]=\dfrac{1}{42}(k+1)(k+2)(2k+3)$$
and
$$(k+1)(2k+3)\ge (k+3)(k+4)$$
so
$$\dfrac{n}{F_{n}}\le \dfrac{42}{(n+1)(n+2)}$$
so
$$\sum_{k=1}^{n}\dfrac{k}{F_{k}}\le 1+2+\dfrac{3}{2}+\dfrac{4}{3}+42\left(\dfrac{1}{6}-\dfrac{1}{n+2}\right)<13$$