I'm trying to prove that the the first derivative of the exponential generating function of the Harmonic numbers, $H_n$, is the exponential generating function of $H_{n+1}$, but I my solution seems flawed:

for $H_n = \sum_0^\infty \frac{1}{x}$

the EGF $H(z)= \sum_{n=0}^\infty H_n \frac{z^n}{n!}$.

\begin{align*} Then\, H'(z) = \frac{dH(z)}{dz} &= \sum_{n=0}^\infty H_nn\frac{z^{n-1}}{n!} \tag{ peform differentiation}\\ &= \sum_{n=0}^\infty H_n\frac{z^{n-1}}{(n-1)!} \tag{cancel $n$ against the $n$ in the $n!$}\\ &= \sum_{i=-1}^\infty H_i \frac{z^i}{i!} \tag { let $i=n-1$, change index}\\ &= \sum_{i=0}^\infty H_{i+1} \frac{z^i}{i!} \tag{can i just change the index from $i=-1$ to $i=0$? why?}\\ &= \text{the EGF of $H_{n+1}$}\\ \end{align*}

Is this correct?

  • 2
    $\begingroup$ $H_n$ is not $\sum_{x=0}^{+\infty}\frac{1}{x}$ (it means nothing) but $\sum_{k=1}^{n}\frac{1}{k}$ and your question has little to do with harmonic numbers, specifically. The (formal) derivative of $\sum_{n\geq 0}\frac{a_n}{n!}x^n$ is $\sum_{n\geq 1}\frac{n a_n}{n!} x^{n-1} = \sum_{n\geq 0}\frac{a_{n+1}}{n!}x^n$, end of story. $\endgroup$ – Jack D'Aurizio Feb 24 '18 at 16:24
  • 1
    $\begingroup$ Work on your notation, then all tasks become easier. E.g. it is incorrect to talk about the "exponential g.f. of $a_n$". It makes sense to talk about the "exponential g.f. of the sequence $\{a_n\}_{n \ge 0}$. Its derivative is the exponential g.f. of the sequence $\{b_n\}_{n \ge 0}$ where $b_n = a_{n+1}$. As @JackDÁurizio pointed out, that's all. $\endgroup$ – Hans Engler Feb 24 '18 at 16:31
  • $\begingroup$ Thank you very much for your comments, @JackD'Aurizio $\endgroup$ – RVC Feb 24 '18 at 16:36
  • $\begingroup$ Thank you very much, @HansEngler! $\endgroup$ – RVC Feb 24 '18 at 16:37

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