I was playing around with some calculations and I noticed that the partial sum of the harmonic series:
$$s_n=\sum_{k=F_n}^{F_{n+1}}\frac 1 k$$
where $F_n$ and $F_{n+1}$ are two consecutive Fibonacci numbers have some interesting properties. It is close to $\frac 1 2$ for small values of $n$ and it seems to converge to a value less than $0.5$ for large $n$. This is what I've got so far:
$$\lim_{n\to\infty} s_n\approx 0.481212$$
I googled a bit to see if there is some theorems or resources for this, and found nothing. I suspect that the series might converge to a smaller number and I may have reached some computational limitations which led to the conclusion that the limit is close to $\frac 1 2$. So my questions are:
- Can we show that the series converge to a non-zero value?
- In case the first answer is yes, can the limit be expressed in a closed form?
Best Answer
In terms of the harmonic numbers $H_n$, your sequence is
$$ s_n = H_{F_{n+1}} - H_{F_n-1} $$
As $n \to \infty$ it's known that $H_n = \log n + \gamma + o(1)$, so
$$ \begin{align} s_n &= \log F_{n+1} + \gamma + o(1) - \log(F_n-1) - \gamma - o(1) \\ &= \log F_{n+1} - \log(F_n-1) + o(1). \end{align} $$
Now $F_m \sim \varphi^m/\sqrt{5}$, where $\varphi$ is the golden ratio, so using the fact that $a \sim b \implies \log a = \log b + o(1)$ we have
$$ \begin{align} s_n &= \log(\varphi^{n+1}/\sqrt{5}) - \log(\varphi^{n}/\sqrt{5}) + o(1) \\ &= \log \varphi + o(1). \end{align} $$
In other words,
$$ \lim_{n \to \infty} \sum_{k=F_n}^{F_{n+1}} \frac{1}{k} = \log \varphi. $$