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:

  1. Can we show that the series converge to a non-zero value?
  2. In case the first answer is yes, can the limit be expressed in a closed form?

3 Answers 3


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,

$$ \lim_{n \to \infty} \sum_{k=F_n}^{F_{n+1}} \frac{1}{k} = \log \varphi. $$

  • $\begingroup$ It might be better to use $s_n\approx ...$ instead of $s_n=...$ $\endgroup$
    – polfosol
    Oct 30, 2016 at 8:14
  • 3
    $\begingroup$ @polfosol, no I disagree. Everything in my answer is rigorous following the definitions of $\sim$ and little-o notation. $\endgroup$ Oct 30, 2016 at 8:14
  • $\begingroup$ @polfosol, see, for example, here for the definitions. $\endgroup$ Oct 30, 2016 at 8:15
  • $\begingroup$ I didn't notice that. Fair enough $\endgroup$
    – polfosol
    Oct 30, 2016 at 8:15
  • $\begingroup$ I will add a comment that ought to have been done : thanks for your very precise and nice answer. $\endgroup$
    – Jean Marie
    Oct 30, 2016 at 8:35

The Fibonacci numbers increase as $\phi^n$ (where $\phi$ is the golden mean $\frac{1+\sqrt{5}}{2}$), and harmonic numbers increase as $\log n$ (i.e., the natural log). Therefore, the difference between the harmonic numbers for successive Fibonacci numbers will approach $\log\phi \approx 0.481211825...$

To expand a bit, the Fibonacci numbers can be expressed as $\frac{\phi^n - (1-\phi)^n}{\sqrt{5}}$. (Try it! The fact that the equation $f(x+2) - f(x+1) - f(x) = 0$ requires a sum of powers of $\phi$ and $1-\phi$ follows from the fact that these are the solutions to the equation $x^2 - x - 1 = 0$, and the coefficients come from f(1) = f(2) = 1.) The second term vanishes, so large Fibonacci numbers can be approximated quite well as $\frac{\phi^n}{\sqrt{5}}$.

Since one definition of the natural logarithm is the integral from 1 to the parameter of the function $t^{-1}$, the harmonic numbers can be approximated as the natural logarithm, and in fact the difference approaches a constant (called $\gamma$, about 0.577). If you're not familiar with integrals, the fact that the harmonic numbers increase as a logarithm is suggested by Oresme's proof that the harmonic series diverges...

$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \cdots > 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{16} + \cdots$$

...and it just so happens that that logarithm is the natural logarithm.

So if you accept that for very large n, the harmonic numbers approach $\log n$, and that the Fibonacci numbers approach $\frac{\phi^n}{\sqrt{5}}$, then you get for two successive...

$$\log\left(\frac{\phi^{n+1}}{\sqrt{5}}\right) - \log\left(\frac{\phi^n}{\sqrt{5}}\right) = \log\left(\frac{\phi^{n+1}}{\phi^n}\right) = \log\phi$$

($\log x - \log y = \log \frac{x}{y}$ is a natural inverse of $\frac{e^x}{e^y} = e^{x-y}$.)

  • $\begingroup$ I will mark this as accepted if you add some more details ;) $\endgroup$
    – polfosol
    Oct 30, 2016 at 8:10
  • $\begingroup$ ...har-r-r-rumph. $\endgroup$
    – user361424
    Oct 30, 2016 at 8:36
  • $\begingroup$ @user361424 very nice answer, a compliment that ought to have been done by the proposer before asking for "more details" ! $\endgroup$
    – Jean Marie
    Oct 30, 2016 at 8:42
  • $\begingroup$ @JeanMarie It seems I am stuck with this $\endgroup$
    – polfosol
    Oct 30, 2016 at 8:44
  • 1
    $\begingroup$ I think this might be the inverse fastest gun in the west problem... (this answer was originally just the first paragraph, and without the parentheticals explaining the golden mean and clarifying the natural log). $\endgroup$
    – user361424
    Oct 30, 2016 at 8:45

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 $\phi$.


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