Sum of inverse of Fibonacci numbers If $F(n)$ is the nth Fibonacci number, How can I prove that:
$$\sum_{i=1}^{\infty} \frac{1}{F(i)}\approx 3.36\, .$$
 A: Since $F(n) \approx \frac{\phi^n}{\sqrt 5}$ for large enough $n$, you may use that as an approximation, which give you a geometric series. For instance, if we use the approximation from the fifth term on, we get
$$
\frac{1}{1} + \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \sum_{n = 4}^\infty \frac{\sqrt 5}{\phi^n} \approx 3.3612
$$
To get a feel for how accurate this is, $F(5) = 5$, while $\frac{\phi^5}{\sqrt 5} \approx 4.96$, which is a relative error of less than $1\%$. The relative error gets smaller and smaller as $n$ grows, by a factor of about $3$, and it is alternating (every other term is too large, every other term is too small) which also reduces the error.
Since every term in the sum is less than $1\%$ away from the corresponding term in the original sequence, the true answer is within $1\%$ of the sum we have, which is $\sum_{n = 4}^\infty \frac{\sqrt 5}{\phi^n} \approx 0.53$. Thus the error of this approximation is at most about $0.005$. One more exact term, and the absolute error will be at most $0.001$.
A: $$
\begin{align}
F_n
&=\frac{\phi^n-(-1/\phi)^n}{\sqrt5}\\
&=\frac{\phi^n}{\sqrt5}\left(1-\left(-\frac1{\phi^2}\right)^n\right)
\end{align}
$$
Therefore,
$$
\begin{align}
\sum_{n=1}^\infty\frac1{F_n}
&=\sum_{n=1}^\infty\frac{\sqrt5}{\phi^n}\left(1+\left(-\frac1{\phi^2}\right)^n+\left(-\frac1{\phi^2}\right)^{2n}+\left(-\frac1{\phi^2}\right)^{3n}+\cdots\right)\\
&=\sqrt5\left(\frac1{\phi-1}-\frac1{\phi^3+1}+\frac1{\phi^5-1}-\cdots\right)\\
&=\sqrt5\sum_{k=0}^\infty\frac{(-1)^k}{\phi^{2k+1}-(-1)^k}
\end{align}
$$
Since $\frac{\sqrt5}{\phi^{19}+1}=0.0002392$,
$$
\sqrt5\sum_{k=0}^8\frac{(-1)^k}{\phi^{2k+1}-(-1)^k}=3.3600587
$$
is less than $0.0002392$ too high.
