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Is there a generating function for

$$\tag{1}\sum_{k\geq 1} H^{(k)}_n x^ k $$

I know that

$$\tag{2}\sum_{k\geq 1} H^{(k)}_n x^n= \frac{\operatorname{Li}_k(x)}{1-x} $$

But notice in (1) the fixed $n$.

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1 Answer 1

up vote 6 down vote accepted

Let $\psi(x)=\frac{\Gamma'}{\Gamma}(x)=\frac{d}{dx}\log\Gamma(x)$ be the digamma function. For $N$ a positive integer, we have $$ \psi(x+N)-\psi(x)=\sum_{j=0}^{N-1}\frac{1}{x+j} $$ (this follows from $x\Gamma(x)=\Gamma(x+1)$ and induction).

Now \begin{eqnarray*} \sum_{k\geq 1}H_n^{(k)}x^k&=&\sum_{k\geq 1}\sum_{j=1}^n \frac{1}{j^k}x^k\\ &=&\sum_{j=1}^n \sum_{k\geq 1} \left(\frac{x}{j}\right)^k\\ &=&\sum_{j=1}^n \frac{x}{j-x}\\ &=&x\sum_{j=1}^n \frac{1}{j-x}\\ &=&x\sum_{j=0}^{n-1}\frac{1}{-x+1+j}\\ &=&x(\psi(-x+1+n)-\psi(-x+1)) \end{eqnarray*}

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