Sum of inverse of Fibonacci numbers If $F(n)$ is the nth Fibonacci number, How can I prove that:
$$\sum_{i=1}^{\infty} \frac{1}{F(i)}\approx 3.36\, .$$

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 $0.001$.
