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The $k^{th}$ harmonic number is given as \begin{equation} H_{k} = \sum_{i=1}^{k} \frac{1}{i} \end{equation}

I am interested in the following series: \begin{equation} \sum_{k=1}^{\infty} \frac{\left(H_{k} \right)^{-k}}{k!} \tag{1} \end{equation}

and for some real positive (possibly integer) $\lambda$: \begin{equation} \sum_{k=1}^{\infty} \frac{\lambda^k \left(H_{k} \right)^{-k}}{k!} \tag{2} \end{equation}

According to Wolfram alpha, $\left( 1 \right)$ is convergent to some expression, which (taking terms in the sum), is something like $1 + \frac{2}{9} + \frac{36}{11^3} + ...$ , while $\left( 2 \right)$ also appears to converge for some $\lambda \in \mathbb{N}$ that I have tried.

What I want for Christmas is to know how to determine the limit of the sum in $\left( 2 \right)$ in terms of $\lambda $, or even an upper bound. As a starting point, I tried to work out the analytic limit of the sum in $\left( 1 \right)$ but failed.

This is similar to other questions 1,2, except with the factorial of $k$ and not quite using the generalised harmonic numbers $\left( H_{k,n} \ne {\left(H_{k}\right)}^{n} \right)$

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  • $\begingroup$ I would not expect a nice closed form. We can deal with fixed powers of the harmonic numbers in the coefficients, hardly we can do the same with $H_k^k$. $\endgroup$ Jun 9, 2015 at 12:52
  • $\begingroup$ Thanks - you are probably right, but I figured I would ask anyway. $\endgroup$
    – JP Janet
    Jun 9, 2015 at 13:01
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    $\begingroup$ Series (2) converges for all real (even complex) $\lambda$. Its sum is bounded in modulus by $e^{|\lambda|}$. $\endgroup$ Jun 9, 2015 at 13:03

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