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Consider the product

$$p_n = \prod_{k=1}^n H_k$$

of $n$ successive harmonic numbers $H_k=\sum_{i=1}^k 1/i$.

The sequence of the $p_n$ is listed in OEIS as A097423/A097424.

I am looking for the asymptotic behaviour of $p_n$ as $n\to\infty$.

My first attemps are based on the asymptotic behaviour of the harmonic number itself

$$H_n \simeq \log(n) + \gamma +\frac{1}{2n}-\frac{1}{12 n^2}+...$$

Defining

$$q_{a,1}(n) = \prod_{k=1}^n \frac{H_k}{\log(k) + \gamma +\frac{1}{2k}}$$
$$q_{a,2}(n) = \prod_{k=1}^n \frac{H_k}{\log(k) + \gamma +\frac{1}{2k}-\frac{1}{12 k^2}}$$

I found numerically that

$$q_{a,1}(10^3)\simeq 0.903394162407764$$ $$q_{a,2}(10^3)\simeq 1.006552015286574$$

Questions

1) Are there closed expressions for the $q_{a,i}(\infty)$ in terms of known constants?

2) What can be said about the asymptotic behaviour of the products

$$r_{a,1}(n) = \prod_{k=1}^n \left( \log(k) + \gamma +\frac{1}{2k}\right)$$
$$r_{a,2}(n) = \prod_{k=1}^n \left(\log(k) + \gamma +\frac{1}{2k}-\frac{1}{12 k^2}\right)$$

Dr. Wolfgang Hintze
  • 12,465

2 Answers2

1

This is not an answer since based on numerical simulations.

If $$p_n= \prod_{k=1}^n H_k$$ it seems that $\log(p_n)$ is almost linear with respect to $n$ (this has been done of $1 \leq n \leq 1000$.

Using data points from $100$ to $10000$ by steps of $10$, a quick and dirty nonlinear regression for $\log(p_n)=a+b \,n^c$ leads to $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & -16.152 & 0.284863 & \{-16.72,-15.59\} \\ b & 1.08280 & 0.002889 & \{1.0770,1.0885\} \\ c & 1.07928 & 0.000372 & \{1.07854,1.0800\} \\ \end{array}$$

For $n=1000$, the regression leads to $\log(p_{1000})=1856.25$ while the exact value is $\approx 1855.48$.

Edit after marty cohen's answer

Reusing the same data set as before, we effectively find for the model $$\log(p_n)=a n \log (\log (n))+\frac{b n}{\log (n)}+c$$ the following results $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 1.00819 & 0.00013 & \{1.00793,1.00845\} \\ b & -0.653283 & 0.00185 & \{-0.656948,-0.649618\} \\ c & 1.64706 & 0.02723 & \{1.59301,1.70112\} \\ \end{array}$$ corresponding to a sum of squares equal to $0.262703$.

Discarding the constant $c$ term, we should get $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 1.00110 & 0.00035 & \{1.00041,1.00179\} \\ b & -0.549559 & 0.00428 & \{-0.558049,-0.541069\} \end{array}$$ corresponding to a sum of squares equal to $10.2736$.

For both case, $|b|\approx \gamma$

Claude Leibovici
  • 260,315
1

The obvious thing is to extract the $\ln(n)$ from the product.

$\begin{array}\\ p_n &= \prod_{k=1}^n H_k\\ &= \prod_{k=1}^n \left( \log(k) + \gamma +f(1/k)\right) \qquad\text{where }f(x) = x/2+O(x^2)\\ &= \prod_{k=2}^n \log(k)\prod_{k=1}^n \left( 1 + (\gamma +f(1/k))/\log(k)\right) \qquad\text{start the first product at 2 otherwise it is zero}\\ &=g(n)h(n)\\ \end{array} $

Note: This estimate changed.

$\begin{array}\\ g(n) &=\prod_{k=2}^n \log(k)\\ \text{so}\\ \log(g(n)) &=\sum_{k=2}^n \log\log(k)\\ &\approx\int_{k=2}^n \log\log(x)dx\\ &= x\log\log(x)|_{2}^n-\int_2^n dx/\ln(x)\\ &\approx n\log\log(n)-n/\log(n)\\ \end{array} $

and $h(n) =\prod_{k=1}^n \left( 1 + (\gamma +f(1/k))/\log(k)\right) $ so

$\begin{array}\\ \ln(h(n)) &=\sum_{k=1}^n \log\left( 1 + (\gamma +f(1/k))/\log(k)\right)\\ &\approx\sum_{k=1}^n (\gamma +f(1/k))/\log(k))\\ &\approx\dfrac{n\gamma}{\log(n)}+O(\log\log(n)) \\ \end{array} $

so, finally, $\log(p_n) \approx n\log\log(n)-n/\log(n)+\dfrac{n\gamma}{\log(n)}+O(\log\log(n)) = n\log\log(n)-(1-\gamma)n/\log(n)+O(\log\log(n)) $.

This could be tested. $\log(g(n))$ could have more terms from a better estimate for $\sum_{k=1}^n \log\log(k)$.

marty cohen
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