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Let $H_n$ is the sum of the reciprocals of the first n natural numbers: $$H_n = \frac{1}{1} +\frac{1}{2}+\frac{1}{3}+ \cdots +\frac{1}{n} $$ $H_n$ is defined only on the natural numbers. Let's think about $H_n$ on real numbers and call the function $H(x)$. Then $H(x)$ might satisfy the recurrence relation $$H(x)-H(x-1)=\frac{1}{x}$$ and $$H(x+a)-H(x)>0\quad(a>0) $$ Then, $H(x)$ is $$H(x)=\left(\frac{1}{1}-\frac{1}{x+1}\right)+\left(\frac{1}{2}-\frac{1}{x+2}\right)+\left(\frac{1}{3}-\frac{1}{x+3}\right)+\cdots \quad(*)$$ I can not prove why $H(x)$ must be a form $(*)$. Please help me to solve the problem.

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  • $\begingroup$ first of all, it is impossible to define it on all the real numbers. Maybe only on positive numbers? $\endgroup$ – Exodd Mar 22 '17 at 12:15
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    $\begingroup$ There are tons of functions $G$, increasing on $\mathbb R_+$, such that $G(x)-G(x-1)=\frac1x$ for every $x>0$ and $G(n)=H_n$ for every natural number $n$, and $(\ast)$ obviously does not hold for all of them. $\endgroup$ – Did Mar 22 '17 at 12:16
  • $\begingroup$ How are we going "to think" of $\;H_n\;$ the reals? What would $\;H(\pi)\;$ be, for example? $\endgroup$ – DonAntonio Mar 22 '17 at 12:20
  • $\begingroup$ Work backwards. $\endgroup$ – Freeman Cheng Mar 22 '17 at 12:27
  • $\begingroup$ @DonAntonio: for instance through $$ H_{\pi}=\int_{0}^{1}\frac{x^\pi-1}{x-1}\,dx$$ $\endgroup$ – Jack D'Aurizio Mar 22 '17 at 17:05
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Suppose there is a function $H : \Bbb R_+ \to \Bbb R$ satisfying :

  1. $\forall x \ge 1, H(x) = H(x-1)+ \frac 1x$
  2. $H$ is increasing
  3. $H(0)=0$ (or $H$ extends the normal harmonic numbers on $\Bbb N$)

Property (2) is very strong, because if you know the exact values of $H(n)$ and $H(n+1)$ (and you do thanks to (3)), then you get for any $x \in [n ; n+1]$, that $H(x) \in [H(n) ; H(n+1)]$, which is (thanks to (1)) an interval of length $\frac 1{n+1}$.
This means that you know $H(x)$ with a precision of about $\frac 1n$.

Meanwhile, if $m$ is an integer and you know $H(x+m)$ with any precision, then (1) gives you an exact number for$H(x+m)-H(x)$, and so you can deduce $H(x)$ with the same precision as $H(x+m)$.

Therefore, combining the two, if say you want to know about $H(x)$ with precision $1/n$ you can simply approximate $H(x+n)$ with $H(m)$ where $m$ is the nearest integer, then use the exact relation between $H(x+n)$ and $H(x)$ to get a good approximation on $H(x)$.

If you look closely, this is exactly what the $(*)$ limit is doing except that it approximates $H(x+n)$ with $H(n)$ instead of using a closer integer (it turns out it still works). Anyway this should convince you that if such an $H$ exists then it is unique:

if $H_1$ and $H_2$ are two solutions then $H_1(x)-H_2(x)$ is a $O(\frac 1x)$ and it is also $1$-periodic, so it is zero forall $x$.


Now I will prove that the series defined in $(*)$ does what you want :

First, the series has a limit, because for any $x \ge 0$, $\frac 1{n}- \frac 1{x+n} = \frac x {n^2+xn} \le \frac x {n^2} = O(\frac 1{n^2})$, which is summable.

So we can define $H(x) = \sum_{k=1}^\infty (\frac 1 k - \frac 1 {x+k})$.

Proving (2) is easy : $H$ is increasing because every term in the series is positive and increasing in $x$.

Proving (3) is also immediate : $H(0) = \sum 0 = 0$

For (1), we have to compute $H(x) - H(x-1)$. To do this we shift the first series by $1$ then subtract termwise :
$H(x) = \sum_{k=1}^\infty (\frac 1 {k} - \frac 1 {x+k}) = \sum_{k=2}^\infty (\frac 1 {k-1} - \frac 1 {x+k-1})$, and
$H(x-1) = \sum_{k=1}^\infty (\frac 1 {k} - \frac 1 {x+k-1}) = (1-\frac 1x) + \sum_{k=2}^\infty (\frac 1 {k} - \frac 1 {x+k-1})$, and so
$H(x)-H(x-1) = (-1+\frac 1x) + \sum_{k=2}^\infty (\frac 1{k-1} - \frac 1k) = -1+\frac 1x + 1 = \frac 1x$

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I will go in the opposite direction took by mercio, hoping to shed more light on the topic.
For any $x\in\mathbb{R}^+$ we may define $$ g(x) = \sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+x}\right) = x \sum_{n\geq 1}\frac{1}{n(n+x)}\tag{1}$$ through an absolutely convergent series, whose formal derivative still is an absolutely convergent series. This grants $g(x)$ is a differentiable function on $\mathbb{R}^+$, and by iterating the argument we also have $g\in C^{\infty}(\mathbb{R}^+)$. If $x=m\in\mathbb{N}^+$ we have that $g(x)$ is defined through a telescopic series:

$$ g(m) = \lim_{N\to +\infty}\sum_{n=1}^{N}\left(\frac{1}{n}-\frac{1}{n+m}\right) = \lim_{N\to +\infty}\left[H_m+(H_N-H_{N+m})\right]=H_m\tag{2}$$ hence the $g$ function provides a $C^\infty(\mathbb{R}^+)$ extension of the function $H_m$, previously defined only for $m\in\mathbb{N}$. Through the Bohr-Mollerup theorem it is also possible to show that $g(m)$ is the only extension of $H_m$ on $\mathbb{R}^+$ that is continuous and increasing, since by the Weierstrass product for the $\Gamma$ function $$ g(m) = \gamma+\psi(m+1) = \gamma + \left.\frac{d}{dx}\log\Gamma(x)\right|_{x=m+1} \tag{3}$$ where $\gamma$ is the Euler-Mascheroni constant, $$ \gamma=\lim_{N\to +\infty}(H_n-\log N) = \sum_{n\geq 1}\left[\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right]\approx 0.5772156649.\tag{4}$$ We may use $(2)$ to compute many explicit values for $H_m$ at non-integer $m$s. For instance, $$H_{\frac{1}{2}}=2\sum_{n\geq 1}\left(\frac{1}{2n}-\frac{1}{(2n+1)}\right)=2\sum_{m\geq 2}\frac{(-1)^{m}}{m}=2-\log(4).\tag{5}$$ We may check that $$ h(m) = \int_{0}^{1}\frac{x^m-1}{x-1}\,dx \tag{6} $$ provides another continuous and increasing extension of $H_m$ on $\mathbb{R}^+$,
hence $H_m = g(m) = h(m)$ for any $m\in\mathbb{R}^+$ and $$ h(m+1)-h(m) = \int_{0}^{1} x^m\,dx = \frac{1}{m+1} \tag{7}$$ is trivial.

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