<s>[INST] 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
 [/INST] Clearly: $F_n > F_{n-1}$, for $n \geq 3$
Thus, $F_n = F_{n-1} + F_{n-2} > 2*F_{n-2} > 2^2*F_{n-4}>...>2^{(n-3)/2}*F_3 = 2^{(n-1)/2}$
Now, your inequality can be written as: $\sum\limits_{i=3}^n \frac{i}{F_i} < 10$
The $LHS < \sum\limits_{i=3}^n \frac{i}{2^{(i-1)/2}} = 2^{1/2}*\sum\limits_{i=3}^n \frac{i}{2^{i/2}} < 2^{1/2}*\sum\limits_{i=3}^\infty \frac{i}{2^{i/2}} = 2^{1/2}*\sum\limits_{i=3}^\infty f(i)$
It suffices to prove that $\sum\limits_{i=3}^\infty f(i) < 5*2^{1/2}$
It's easy to verify that $f(i)$ is non-negative, monotone decreasing on $[3;\infty)$, thus integral test is applicable and we have:
$\sum\limits_{i=3}^\infty f(i) \leq f(3)+ \int_{3}^{\infty} f(x)dx = \frac{3}{2^{3/2}}+ \frac{1+3*\ln a}{(\ln a)^2}*a^{-3} = 7.06457 < 7.0710 = 5*2^{1/2}$, where $a=2^{1/2}$.
This upper approximation is far worse than that of Simon Rose.
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