Harmonic Number for $H_{n+1}$ and $H_{n+\frac{1}{2}}$

Question

According to the definition of harmonic number $H_n = \sum\limits_{k=1}^n\frac{1}{k}$.

How we can define $H_{n+1}$ and $H_{n+\frac{1}{2}}$?

Answer 1

$$H_{n+1} = \sum_{k=1}^{n+1} \dfrac{1}{k} = H_n + \frac{1}{n+1}$$

$H_{n+1/2}$ is not defined by the formula for $H_n$, unless you want to just say it's the same as $H_n$. However, one function whose restriction to the positive integers is $H_n$ is $\psi(n+1)+\gamma$, where $\psi$ is the digamma function and $\gamma$ is the Euler-Mascheroni constant. This would make

$$ H_{n+1/2} = \psi(n+3/2)+ \gamma$$

EDIT: Since $$\psi(n+3/2) = 2 \psi(2n+2) - \psi(n+1) - 2 \ln 2$$

you could also write $$ H_{n+1/2} = 2 H_{2n+1} - H_n - 2 \ln 2 $$

Answer 2

Note that $$ H_n=\sum_{k=1}^n\frac{1}{k}=\int_0^1\frac{1-x^n}{1-x}dx. $$ Hence $$ H_{n+1}=\int_0^1\frac{1-x^{n+1}}{1-x}dx $$ and $$ H_{n+1/2}=\int_0^1\frac{1-x^{n+1/2}}{1-x}dx. $$

Answer 3

Well, the question is more or less the same as how do we define $n!$ for $n\not\in\mathbb{N}?$
Intro: there are many functions agreeing with $n!$ over $\mathbb{N}$, but it we impose that

1. $f(1)=1$ and $f(x+1)=x\,f(x)$ for any $x\in\mathbb{R}^+$
2. $\log f$ is convex over $\mathbb{R}^+$

then we may prove that the $\Gamma$ function, defined through $$ \Gamma(n+1)=\int_{0}^{+\infty} x^n e^{-x}\,dx $$ is the only natural extension of the factorial function. This is the Bohr-Mollerup theorem.
(Have a look at the section about special functions in my notes, too). The integral representation and the property $\Gamma(x+1)=x\,\Gamma(x)$ provide an analytic continuation to the complex plane: $\Gamma(x)$ turns out to be a meromorphic function with simple poles at $0,-1,-2,-3,\ldots$ and residues $\frac{1}{0!},-\frac{1}{1!},\frac{1}{2!},-\frac{1}{3!},\ldots$ Weierstrass factorization theorem then gives $$\Gamma(z+1) = e^{-\gamma z}\prod_{n\geq 1}\left(1+\frac{z}{n}\right)^{-1}e^{z/n} $$ where $\gamma=\lim_{n\to +\infty}\left(H_n-\log n\right)$. Here we go: by applying $\frac{d}{dz}\log(\cdot)$ to both sides of the previous identity, and by defining $\psi(z)$ as $\frac{\Gamma'(z)}{\Gamma(z)}$, we have $$ \psi(z+1) = -\gamma+\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+z}\right) $$ so the most natural extension of $H_n$ is given by $$ H_n = \gamma+\psi(n+1)=\gamma+\int_{0}^{+\infty}\sum_{m\geq 1}\left(e^{-mu}-e^{-(m+n)u}\right)\,du =\gamma+\int_{0}^{+\infty}\frac{1-e^{-nu}}{e^u-1}\,du.$$

The duplication and reflection formulas for the $\Gamma$ function give duplication and reflection formulas for the $\psi$ function, too, such that

$$ H_{n+1/2} = 2H_{2n+1}-H_n-2\log 2.$$ That can be proved from the integral representation only, if needed. The ultimate tool is Gauss' Digamma Theorem, essentially stating that $\psi(x)$ (so $H_{x-1}$ too) has a not-so-involved closed form for any $x\in\mathbb{Q}^+$.