TITLE: Partial sum of the harmonic series between two consecutive fibonacci numbers
QUESTION [12 upvotes]: 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?

REPLY [3 votes]: Way late to the party, but here's  a general result, and an elementary derivation:

Claim: Let $(a_n)$ and $(b_n)$ be sequences of positive integers with $a_n\to\infty$ and $\lim_{n\to\infty}\frac{b_n}{a_n}=c$. Then
$$
\lim_{n\to\infty}\sum_{k=a_n}^{b_n}\frac1k=\log c.$$
Proof: Start with the inequalities
$$\frac{x-1}x\le\log x\le x-1.$$ Substitute $x=(k+1)/k$ into the right inequality and $x=k/(k-1)$ into the left, obtaining
$$\log(k+1)-\log k\le\frac1k\le\log k-\log(k-1).$$ Sum from $k=a_n$ to $k=b_n$, and use telescoping to find
$$
\log\frac{b_n+1}{a_n}\le\sum_{a_n}^{b_n}\frac1k\le\log\frac{b_n}{a_n-1}.$$ Finally, take the limit as $n\to\infty$.

Now apply this result with $a_n=F_n$ and $b_n=F_{n+1}$ and use the fact that  $F_{n+1}/F_n$ tends to the golden ratio $\phi$.