# Proof of an exact formula for $H_n$

The $$n$$th harmonic number $$H_n$$ is defined as $$H_n=\sum_{n\geq k\geq 1}\frac{1}{k}$$ A good approximation for this is $$H_n\approx \gamma+\log n +\frac{1}{2n}$$ Where $$\gamma$$ is the Euler-Mascheroni constant.
In the book Table of integrals, infinite series, and products by I.S Gradshteyn and A.M Ryzhik, it is given that $$H_n=\gamma+\log n+\frac{1}{2n}+\sum_{k\geq 2}\frac{A_n}{n(n+1)...(n+k-1)}$$ where $$A_k=\frac{1}{k}\int_{0}^{1}x(1-x)(2-x)...(k-1-x)dx$$ This is an exact formula. I found it pretty remarkable. How can one prove this?
There is another question. The starting values of $$A_n$$ are $$0,\frac{1}{12},\frac{1}{12},\frac{9}{20}$$. Is there any pattern in these numbers? Another way to say this it: is there any closed form for the integral?
Update: I found in Wolfram mathworld that $$H_n=\gamma+\psi_{0}(n+1)$$ Where $$\psi_{0}(x)$$ is the digamma function. I did a little research and found $$\psi_{0}(1+z)=\log(z)+\frac{1}{2z}-\sum_{j\geq 1}\frac{B_{2j}}{2jz^{2j}}$$ so $$H_n=\gamma+\log(n)+\frac{1}{2n}-\sum_{j\geq 1}\frac{B_{2j}}{2jz^{2j}}$$ so the only thing that is left to prove is $$\sum_{j\geq 1}\frac{B_{2j}}{2jn^{2j}}=-\sum_{k\geq 2}\frac{A_n}{n(n+1)...(n+k-1)}$$ where $$B_{2n}$$ are the Bernoulli numbers.
How can we prove the equality of those two series? From this the original formula for $$H_n$$ can be proved. You can give the proof in any way you like but the proof of the equality of these two series or the proof with the fact that $$H_n=\gamma+\psi_{0}(n+1)$$ would be the best.
Update: From the wikipedia page of Gregory coefficients (link to this article was given by Donald Splutterwit) I found some properties of $$A_k$$ from the references of that wikipedia article. This is the article about $$A_k$$. They are written as $$P_{n+1}(y)$$ in this paper.

• $A_k$ might be related to ... en.wikipedia.org/wiki/… Commented Oct 18, 2020 at 12:54
• $H_n=\sum_{n\geq k\geq 1}\frac{1}{k}$ is an exact formula, too. It's just a finite sum, not a cool infinite one, but... you can't have all.
– user436658
Commented Nov 11, 2020 at 20:54

Note that by the Beta function formula we have $$\int_0^1 t^{k-1}(1-t)^{n-1}\,dt = \frac{\Gamma(k)\Gamma(n)}{\Gamma(n+k)} = \frac{(k-1)!}{n(n+1) \cdots (n+k-1)}.$$
Therefore, the series \begin{align} \sum\limits_{k=1}^{\infty} \frac{A_k}{n(n+1) \cdots (n+k-1)} &= \sum\limits_{k=1}^{\infty} \int_0^1 \frac{A_k}{(k-1)!}t^{k-1}(1-t)^{n-1}\,dt \tag{1} \\&= \sum\limits_{k=1}^{\infty} \int_0^1\int_0^1 (-1)^{k-1}\frac{\Gamma(x+1)}{k!\Gamma(x+1-k)}t^{k-1}(1-t)^{n-1}\,dx\,dt \tag{2} \\&= \int_0^1\int_0^1 \left[\sum\limits_{k=1}^{\infty}(-1)^{k-1}\binom{x}{k}t^{k-1}\right](1-t)^{n-1}\,dx\,dt \tag{3} \\&= \int_0^1\int_0^1 \left[\frac{1 - (1-t)^x}{t}\right](1-t)^{n-1}\,dx\,dt \tag{4} \\&= \int_0^1\int_0^1 \left[\frac{1 - t^x}{1-t}\right]t^{n-1}\,dx\,dt \tag{5} \\&= \int_0^1 \left[\frac{1}{1-t} + \frac{1}{\ln t}\right]t^{n-1}\,dt \tag{6} \end{align} where, in line $$(2)$$ we used the expression for Gamma function and in line $$(3)$$ we used the generalized Binomial theorem.
The integral $$(6)$$ is $$\displaystyle \int_0^1 \left[\frac{1}{1-t} + \frac{1}{\ln t}\right]t^{n-1}\,dt = H_n - \ln n - \gamma$$, a standard Cauchy-Frullani integral.