In short, my question is asking to prove that the $$\lim_{n\to\infty}\frac{\text{number of digits in the denominator of} \sum_{k=1}^{10^n} \frac 1k}{10^n}=\log_{10} e$$ I know that the number of digits in a number is $\lfloor \log_{10} n\rfloor +1$ and that the Harmonic numbers are given by $\gamma+\psi_0 (n+1)$, where $\psi_0 (x)=\frac{\Gamma'(x)}{\Gamma(x)}$ but I don't see how to find the number of digits in the denominator of $\psi_0(n+1)+\gamma$.

Context:From Wolfram MathWorld,

"The numbers of digits in the denominator of $H_{10^n}$ for $n=0, 1, \ldots$ is given by $1, 4, 40, 433, 4345, 43450, 434110, 4342302, 43428678, \ldots$ (OEIS A114468). These digits converge to what appears to be the decimal digits of $\log_{10}e=0.43429448\ldots$ (OEIS A002285)."

  • $\begingroup$ "converge to what appears to be..." That doesn't sound like a definite statement of mathematical fact. Are you sure this isn't an open question? $\endgroup$ May 21, 2015 at 0:00
  • $\begingroup$ @user2566092 I'm not sure if it is or isn't, but the OEIS link does not say that it is an open question. Also, open questions are not banned? $\endgroup$
    – Teoc
    May 21, 2015 at 0:03
  • $\begingroup$ @Vladmir No of course open questions aren't banned, especially because there are some questions that were probably answered before but have been largely "lost" or forgotten now. I've asked a few questions like that before here and often gotten positive answers. However, if OEIS makes it sound like an open question, that some probably profressional mathematician considered before making the OEIS post, then it may be very difficult to answer and better suited for MathOverflow. That's where people usually post professional mathematics research level questions. $\endgroup$ May 21, 2015 at 0:07

2 Answers 2


The denominator of the $M$th harmonic number, before cancelling out any common factors with the numerator, is $\mathop{\rm lcm}[1,\dots,M]$; by the prime number theorem, we know that $\ln \mathop{\rm lcm}[1,\dots,M] \sim M$, and hence the number of digits of $\mathop{\rm lcm}[1,\dots,M]$ is approximately $M/\ln10 = M\log_{10}e$.

The remaining step would be to show that there isn't too much cancellation between the numerator and denominator of $M$. I'm willing to believe that that is an open problem, though I don't know for sure.

  • $\begingroup$ How does the PNT, which states that $\pi(x)\approx\frac{x}{\log x}$ and $p_n\approx n\log n$ imply $\log \lcm[1\ldots M]\approx M$? $\endgroup$
    – Teoc
    May 27, 2015 at 0:27
  • 1
    $\begingroup$ The function $\log \mathop{\rm lcm}[1,\dots,x]$ is typically denoted $\psi(x)$; most proofs of the prime number theorem start by establishing that $\psi(x) \sim x$ and then deducing that $\pi(x) \sim x/\log x$. See en.wikipedia.org/wiki/Prime_number_theorem#Proof_sketch to start. $\endgroup$ May 27, 2015 at 1:07

Let $D(n)$ be the denominator of $H(n)$ and let $L(n) = LCM(1,2,\ldots,n)$. It is easy enough to give bounds good enough to show that $$\lim_{n \to \infty} \frac{\log L(n)}{n} = \lim_{n \to \infty} \frac{\log D(n)}{n} = 1.$$ (All logs are base $e$.)

It is well known that $\lim_{n \to \infty} \frac{\log L(n)}{n}=1$, and clearly $D(n) \leq L(n)$, so the point is to provide lower bounds. Fix a large positive integer $k$. Choose $N$ large enough that $N>k^2$ and $N/k$ is large than any prime dividing the numerator of one of $H_1$, $H_2$, ..., $H_k$. We will bound $H_n$ for $n>N$.

Let $p$ be a prime between $n/k$ and $n$. Then the summands in $H_n$ whose denominator is divisble by $p$ are $1/p$, $1/(2p)$, $1/(3p)$, ..., $1/(\ell p)$ for some $\ell \leq k$. They add up to $H_{\ell}/p$. Since $p>n/k$ and $\ell \leq k$, we know that $p$ does not divide the numerator of $H_{\ell}$. Set $H_{\ell} = a/b$. Then $H_n$ is of the form $(a/b)/p + c/d=(ad+pcb)/(pbd)$ where $p$ does not divide $a$ or $d$. So $p$ divides the denominator of $H_n$.

We see that $$\log D(n) \geq \sum_{n/k<p<n} \log p = \vartheta(n) - \vartheta(n/k) \sim n\cdot (1-1/k).$$ Here $\vartheta(x)$ is Chebyshev's notation for $\sum_{p < x} \log p$ and the asymptotic relation is from the prime number theorem. So $\lim \inf_{n \to \infty} \frac{\log D(n)}{n} \geq 1-1/k$. Since $k$ was arbitrary, $\lim \inf_{n \to \infty} \frac{\log D(n)}{n} \geq 1$.

Examples suggest that $L(n)/D(n)$ is very small compared to $L(n)$: You can see some values here.


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