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.

  • $\begingroup$ 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
  • 3
    $\begingroup$ 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

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
$H(x)-H(x-1) = (-1+\frac 1x) + \sum_{k=2}^\infty (\frac 1{k-1} - \frac 1k) = -1+\frac 1x + 1 = \frac 1x$


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.