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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?
polfosol
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3 Answers3

22

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,

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

Antonio Vargas
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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 $\frac{e^x}{e^y} = e^{x-y}$.)

user361424
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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 $\phi$.

grand_chat
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