For positive integers $s$ and $n$ (let's limit the generality), define $$H_s(n)=\sum_{k=1}^{n}\frac{1}{k^s},\qquad G_s(n)=\sum_{k=1}^{n}\binom{n}{k}\frac{(-1)^{k-1}}{k^s}.$$

The former is well-known; the latter is what I'm wondering about. I met it in a number of contexts. Say, if $X_1,\ldots,X_n$ are independent random variables distributed exponentially with parameter $1$, then $s!G_s(n)$ is the $s$-th moment of $\max\{X_1,\ldots,X_n\}$.

I was looking for an expression of $G_s(n)$ in terms of $H_s(n)$. Here are examples: \begin{align} G_1(n) &= H_1(n), \\ G_2(n) &= (H_2(n)+H_1^2(n))/2, \\ G_3(n) &= (2H_3(n)+3H_2(n)H_1(n)+H_1^3(n))/6, \\ G_4(n) &= (6H_4(n)+8H_3(n)H_1(n)+3H_2^2(n)+6H_2(n)H_1^2(n)+H_1^4(n))/24 \end{align} (the first one is seen in the linked article and in some questions on this site).

The general formula appears to be

$$G_s(n)=\sum_{\substack{a_1,\ldots,a_s\geqslant 0\\ 1\cdot a_1+\ldots+s\cdot a_s=s}}\prod_{k=1}^{s}\frac{H_k^{a_k}(n)}{k^{a_k}\cdot a_k!}.$$

How would one prove it best? Can it be seen from $$\sum_{s=0}^{\infty}G_s(n)t^s=\prod_{k=1}^{n}\Big(1-\frac{t}{k}\Big)^{-1}\tag{$*$}\label{helpeq}$$ which is easy to prove (here $G_0(n)=1$)?

  • 1
    $\begingroup$ Yeah, it can be seen from your identity. Hint: $\prod_{k=1}^n \left(1-\dfrac{t}{k}\right)^{-1} = \exp\left(\sum_{i \geq 1} H_i\left(n\right) t^i / i\right)$. Then, recall a formula for how to write the coefficients of $\exp f$ in terms of the coefficients of $f$, where $f$ is an arbitrary formal power series with constant term $0$. $\endgroup$ May 19, 2019 at 18:37
  • $\begingroup$ @darijgrinberg: Thanks a lot! (I should have remembered the pattern...) Essentially this answers the question completely. $\endgroup$
    – metamorphy
    May 19, 2019 at 18:48
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    $\begingroup$ @metamorphy The coefficients of the expansion of $G_i$ in terms of $H_i$ are listed in oeis.org/A036039 Irregular triangle of multinomial coefficients of integer partitions read by rows (in Abramowitz and Stegun ordering) giving the coefficients of the cycle index polynomials for the symmetric groups S_n. $\endgroup$ Apr 22, 2022 at 7:38

1 Answer 1


Indeed (thanks to darij grinberg; writing it out just to have the question answered), $$\sum_{s=1}^{\infty}\frac{H_s(n)}{s}t^s=\sum_{k=1}^{n}\sum_{s=1}^{\infty}\frac{1}{s}\Big(\frac{t}{k}\Big)^s=-\ln\prod_{k=1}^{n}\Big(1-\frac{t}{k}\Big)$$ so the result follows from $(*)$ and "exponential formula". In turn, $(*)$ is a consequence of $$\prod_{k=1}^{n}\Big(1-\frac{t}{k}\Big)^{-1}=\sum_{k=1}^{n}(-1)^{k-1}\binom{n}{k}\Big(1-\frac{t}{k}\Big)^{-1}$$ which is simply the partial fraction expansion of the LHS.


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