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For a given sequence of real numbers $(x_n)_{n=1}^\infty \subset [0,1]$, let $A(a,b,N)$ to be the number of terms $x_n$ of the sequence up to index $N$ such that $a \leq x_n \leq b$. A sequence of real numbers $(x_n)_{n=1}^\infty$ is said to be uniformly distributed modulo 1 if $\displaystyle \frac{A(a,b,N)}{N}$ tends to $b-a$ as $N \rightarrow \infty$ for every $0 \leq a < b \leq 1$.

My question concerns whether or not the following sequence is uniformly distributed. Define $x_n = H_n - \lfloor H_n \rfloor$, where $H_n = \displaystyle \sum_{k=1}^n \frac{1}{k}$ is the $n$-th harmonic number. Then is $(x_n)_{n=1}^\infty$ uniformly distributed?

What if we instead defined $x_n$ to be a random element of the interval $[H_{n-1}, H_n]$ with respect to a uniform distribution, and ask if the random sequence $(x_n - \lfloor x_n \rfloor)_{n=1}^\infty$ is uniformly distributed?

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This seems approachable via Weyl's criterion; have you tried this? –  Daniel Litt Mar 31 '11 at 17:09
    
A quibble on terminology. A sequence of real numbers is u.d. mod 1 means the sequence of fractional parts is u.d. - if the numbers are already in $[0,1)$ there is no need to say "mod 1." –  Gerry Myerson Mar 31 '11 at 23:44

2 Answers 2

up vote 8 down vote accepted

$H_n$ is a asymptotic to $\log n+\gamma+O(1/n)$. This means that the values of $x_n$ for $e^k \leq n\leq e^{k+1/2}$ all fall into the same interval of length about $1/2$. The sequence is not equidistributed (the proportion of $x_n$ in that interval for $n\leq e^k$ and for $n\leq e^{k+1/2}$ differ much).

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Indeed, for the same reason the sequence of fractional parts of $\log n$ is not uniformly distributed. It just grows too slowly. –  Gerry Myerson Mar 31 '11 at 23:41

The numbers $e^{H_N-\lfloor H_n\rfloor}$ all fall in the interval $[1,e]$. They do not have a limiting distribution but, in a sense which I will try to make precise, with some smoothing they probably have the uniform distribution in that interval. Otherwise put, the "best" distribution to associate with the given problem assigns the interval $[a,b]$ the probability $\frac{e^b-e^a}{e-1}.$

Let $r(a,b,N)= \frac{A(a,b,N)}{N}.$ It turns out that $$\limsup r(a,b,N) =\frac{e^{1-b}(e^b-e^a)}{e-1} \text{ and }\liminf r(a,b,N)=\frac{e^{-a}(e^b-e^a)}{e-1}.$$ So if $b-a$ is small, then the ratio is close to $e$.

The value of $H_n$ is quite close to $\ln{n}+\gamma$. where $\gamma = 0.577\cdots$ is the Euler Mascheroni constant. As $N$ increases and the values $H_N$ go to infinity at an ever decreasing rate, the count $A(a,b,N)$ stays constant for long stretches during which the ratio $r(a,b,N)$ decreases. For each positive integer $K$ there is a local minimum of $r(a,b,N)$ just before $H_N \gt K+a.$ When that occurs $N \approx e^{K+a-\gamma}.$ Then for a while, every time $N$ increases by $1$ so does $A(a,b,N)$ as well as $r(a,b,N).$ This continues until $H_N \gt K+b,$ which is at $N \approx e^{K+b- \gamma}.$ That is when $r(a,b,N)$ has a local maximum. Finding what those local extreme values are involves summing finite geometric progressions. I did this and compared the results to numerical data ($K=7,8,9$) with which it agrees well.

Notes:

  • The same upper and lower limits occur (and are easier) if one replaces $H_n$ by $\int_1^n\frac{1}{x}dx=\ln{n}$.
  • Until $\frac{1}{N}$ is smaller than $b-a,$ we might have $H_N \lt K+a \lt K+b \lt H_{N+1}.$ Until it is "much" smaller, things might be less regular. This has very little effect on the convergence.
  • The variation where we replace $H_N$ with a random number between $H_N$ and $H_{N+1}$ would be the same.

There are various transformations which take every convergent sequence to another with the same limit but faster convergence. When applied to a divergent but bounded sequence they may take it to another sequence with a limit (and various transformations should all give the same limit). A standard transformation is to replace $\lim r(a,b,N)$ with the limit of an average $$\lim \frac{r(a,b,1)+r(a,b,2)+\cdots + r(a,b,N)}{N}$$ I suspect that that might turn out to have limit $\frac{(e^b-e^a)}{e-1}.$ It might not be hard, but I have not checked it carefully.

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