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Find an exponential or ordinary generating function of reciprocal Harmonic numbers.

$f(x)=\sum\limits_{n=1}^{\infty} \frac{1}{H_n}\frac{x^n}{n!}$ or

$f(x)=\sum\limits_{n=1}^{\infty} \frac{1}{H_n}x^n$

Also, it would be nice to see EGF or OGF for other reciprocals of common "numbers", like binomial coefficients, Stirling numbers, Catalan numbers, etc.

P.S. I suspect one could use Digamma function here

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This is not a complete answer, just too long for a comment.

Let us work with EGF $f(x)=\sum_{n=1}^{\infty} \frac{1}{H_n}\frac{x^n}{n!}$.

Clearly, $$H_n = \int_0^1 \frac{1-x^n}{1-x}dx.$$

Therefore, $$ e^z -1 = \sum_{n=1}^\infty \frac{H_n z^n }{H_n n!} = \int_0^1 \sum_{n=1}^\infty \frac{ z^n (1-x^{n})}{H_n n!}\frac{dx}{1-x} = \\ = \int_0^1 \frac{f(z) - f(zx)}{1-x} dx = z\int_0^1 \int_{x}^1\frac{f'(zu)}{1-x}du\, dx = -z\int_0^1 f'(zu) \log(1-u) du =\\ = -\int_0^z f'(x) \log(1-x/z) dx. $$ This gives some integral equation for $f'$.

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  • $\begingroup$ maybe it would be easier to find a derivation of OGF i.e. $f'(x)=\sum\limits_{n=1}^{\infty} \frac{n}{H_n}x^{n-1}$. By eyeballing the look of that function I got a pretty good match whose series expansion around 0 gives pretty close coeficients to $n/H_n$. Could someone explain why is that? Is it pure luck? Plot[{Sum[n/HarmonicNumber[n] x^(n - 1), {n, 1, 100}], (Coth[-x +1]/Sqrt[(1 - x)])}, {x, -1, 1}] $\endgroup$ – nik Jan 5 '16 at 20:21
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    $\begingroup$ For OGF, the equation is the same, but lhs will be just $z/(1-z)$. And yes, this is a coincidence: the singularity analysis says that the function should behave like $\log(1-x)$ for $x$ close to $1$. $\endgroup$ – zhoraster Jan 5 '16 at 20:34
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For the reciprocals of harmonic numbers, I doubt there is a closed-form generating function.

For the reciprocals of the Catalan numbers, Maple gives me the ogf

$$\dfrac{16+2x}{(4-x)^2} + \dfrac{24 \sqrt{x} \arcsin(\sqrt{x}/2)}{(4-x)^{5/2}}$$

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