Infinite Sum of the Reciprocal of the Inverse Harmonic Number Function

Question

It's a very strange question I know, but I am trying to figure out the infinite series $\sum_{k=1}^\infty \frac {1}{H^{(-1)}_k}$ where $H^{(-1)}_k$ is the inverse of the harmonic number function generalized to all reals. I tried to put it into Wolfram Alpha, but the only way I could figure out to express $H^{(-1)}_k$ was in terms of the $\psi^{(-1)}$ function which they don't have.

Answer 1

One "generalization" is to let $F(x)$ be the least positive integer $n$ for which $H_n \ge x$. Now $H_n = \ln(n) + \gamma + 1/(2n) + O(1/n^2)$, so $F(x) \approx \exp(x - \gamma - 1/(2x) + O(1/x^2))$. In particular $\sum_{k=1}^\infty 1/F(k)$ converges. Of course I don't know exactly what the sum is. But from using the exact values for $k = 1$ to $7$ and numerical approximation from there on, I estimate it to be about $1.392189$.

Answer 2

$\dfrac{H_{k+1}^{-1}}{H_k^{-1}}\approx3$, so the series decreases exponentially, and therefore converges. By using Mathematica's

FindRoot function, the numerical value $S=1.419878024238479212425700930495898499^+$ was obtained, for which the Inverse Symbolic Calculator and Online Encyclopedia of Integer Sequences are unable to return anything. Hope this helps.