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This question is somewhat related to one of my previous questions: Fibonorial of a fractional or complex argument.

Recall the definition of harmonic numbers: $$H_n=\sum_{k=1}^n\frac1k=1+\frac12+\,...\,+\frac1n\tag1$$ Obviously, harmonic numbers satisfy the following functional equation: $$H_n-H_{n-1}=\frac1n\tag2$$ The definition $(1)$ is valid only for $n\in\mathbb N$, but it can be generalized to all positive indices. There are several equivalent ways to do this: $$H_a=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+a}\right)=\int_0^1\frac{1-x^a}{1-x}\,dx=\frac{\Gamma'(a+1)}{\Gamma(a+1)}+\gamma\tag3$$ This generalized definition gives a real-analytic function (that can be extended to a complex-analytic if needed) and still satisfies the functional equation $(2)$ even for non-integer values of $a$.

Now, consider the product of harmonic numbers: $$P_n=\prod_{k=1}^nH_k=H_1\,H_2\,...H_n=1\times\left(1+\frac12\right)\times\,...\times\left(1+\frac12+\,...+\frac1n\right)\tag4$$ The numerators and denominators of the terms of this sequence appear as A097423 and A097424 in the OEIS. Obviously, the following function equations hold: $$\frac{P_n}{P_{n-1}}=H_n,\quad\quad\frac{P_n}{P_{n-1}}-\frac{P_{n-1}}{P_{n-2}}=\frac1n\tag5$$ I'm looking for a continuous generalization $P_a$ of the discrete sequence $P_n$, which is real-analytic for all $a>0$ and satisfies the functional equations $(5)$.

Could you suggest a way to construct such a function? Is there a series or integral representation for it? Can we generalize it to complex arguments?

Update: It seems we can use the same trick that is used to define $\Gamma$-function using a limit involving factorials of integers: $$P_a=\lim_{n\to\infty}\left[\left(H_n\right)^a\cdot\prod_{k=1}^n\frac{H_k}{H_{a+k}}\right]=\frac1{H_{a+1}}\cdot\prod_{n=1}^\infty\frac{\left(H_{n+1}\right)^{a+1}}{\left(H_n\right)^a\,H_{a+n+1}}\tag6$$

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Vladimir Reshetnikov
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  • so with $f(z) =\frac{P(z)}{P(z-1)}$ you want $f(z) - f(z-1) = \frac{1}{z}$ ? – reuns Sep 06 '16 at 01:31
  • Yes, exactly.${}$ – Vladimir Reshetnikov Sep 06 '16 at 02:11
  • I'm also interested in an asymptotic behavior of $P_a$ for $a\to\infty$. – Vladimir Reshetnikov Sep 07 '16 at 03:35
  • In $f(z)-f(z-1) = g(z)$, did you try equating the coefficients ? Here $f(z+a)-f(z+a-1) = \frac{1}{z+a}$, $f(z+a) = \sum_{n=0}^\infty c_n z^n, f(z+a-1) = \sum_{n=0}^\infty c_n (z-1)^n =\sum_{n=0}^\infty c_n \sum_{k=0}^n {n \choose k} (-1)^{n-k} z^{k}$ $ = \sum_{n=0}^\infty z^n \sum_{k=n}^\infty {m \choose n} c_m , \frac{1}{z+a} = a^{-1} \sum_{n=0}^\infty (-1/a)^{n} z^n$ so that $c_n + \sum_{k=n}^\infty {m \choose n} c_m = (-1)^n a^{-n-1}$ – reuns Sep 07 '16 at 03:50
  • For my own exercises I once tried to find a summation formula for the sums of like powers of logarithms $\Lambda_m(a,b)=\log(a)^m+\log(a+1)^m+...+\log(b-1)^m$ in an analoguous fashion as Faulhaber and Bernoulli looked at the sums of consecutive like powers, resulting in the well known polynomials. Those polynomials allow also to generalize the summation-bounds to fractional or general numbers. I developed a formal power series for that sums-of-(powers of) logarithms by the technique of indefinite summation. Surprisingly (or not?) that was exactly ... – Gottfried Helms Sep 07 '16 at 23:31
  • the power series for the log of the gamma (lngamma in Pari/GP) and provides thus a path to the interpolation to fractional arguments of the factorial function, just having rediscovered Euler's solution. So possibly such an ansatz can also help here; unfortunately we have to do with the digamma-function (psi-function in Pari/GP) here for which I do not yet have an idea how to apply the indefinite summation for the logarithm of your $P_a$-function. I'm looking at some re-engineering of the psi-function to have it better manageable and invertible but have not yet really succeeded so far... – Gottfried Helms Sep 07 '16 at 23:36
  • @GottfriedHelms Your story reminds me of my another question. – Vladimir Reshetnikov Sep 07 '16 at 23:43
  • So it's obviously a corpus of much related questions, having a common core-focus... If it is of interest, I could also post a link to my results for the logarithm-sums at the other question? (It is at http://go.helms-net.de/math/divers/BernoulliForLogSums.pdf ) – Gottfried Helms Sep 08 '16 at 01:00
  • @GottfriedHelms Thanks, it is an interesting reading. My motivation for the question about powers of logarithm summation was (still is) an attempt to generalize Stieltjes constants to fractional (or even complex) indexes. – Vladimir Reshetnikov Sep 08 '16 at 01:02

1 Answers1

1

One may use the Euler-Maclaurin formula to find that as $n\to\infty$, we have

$$\sum_{k=1}^n\ln(H_k)\sim c+\int_0^n\ln(H_x)\ dx+\frac{\ln(H_n)}2+\frac{H_n'}{4H_n}+\dots$$

where $c=\lim_{n\to\infty}\sum_{k=1}^n\ln(H_k)-\left(\int_0^n\ln(H_x)\ dx+\frac{\ln(H_n)}2+\frac{H_n'}{4H_n}\right)\approx0.7288$

One can then use $(5)$ to recursively relate back to the original sum in the same manner you did in $(6)$, as this expansion approximates best as $n\to\infty$.

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