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Inspired by these three questions, I asked myself whether

$$\sum_{n>0}\Big[~H(e^n)-n-\gamma~\Big]~=~0.278091975548622251874828828459627630\ldots$$

might also possesses a closed form expression, where $H(k)$ represents the $k^{th}$ harmonic number, whose generalization to non-natural arguments can be found here. Unfortunately, I personally was not thus far able to adapt the various ingenious methods presented in their answers, so as to make them suit my purpose. Any insightful foray or meaningful approach to help tackle this rather resilient series would therefore be deeply appreciated. Thank you !

Lucian
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  • I doubt you can simplify $\sum_{n \ge 0} \int_0^1(\frac{1-x^{e^n}}{1-x}-n-\gamma)dx$ – reuns Oct 26 '17 at 02:33
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    @reuns: Appearances can sometimes be quite deceiving. – Lucian Oct 26 '17 at 02:37
  • @reuns: I envy your sense of perception. – Lucian Oct 26 '17 at 05:09
  • Honnestly it is obvious your linked limit simplifies. In your series, if you replace $e^n,n$ by $x^n, n \log x$ and if you use a formulation as a series with (almost only) positive terms then you can find the asymptotic as $x \to 1$. – reuns Oct 26 '17 at 05:22

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From the asymptotic expansion of the digamma function $$ \psi(x+1) \approx \log(x)+\frac{1}{2x}-\sum_{m=1}^{M}\frac{B_{2m}}{2m x^{2m}} \tag{A}$$ we get: $$\mathcal{S}= \sum_{n\geq 1}\left[\psi(e^n+1)-n\right]\approx \frac{1}{2(e-1)}-\sum_{m=1}^{M}\frac{B_{2m}}{2m (e^{2m}-1)} \tag{B}$$ but the last series is not convergent as $M\to +\infty$ since Bernoulli numbers have a rapid growth from some point on (their exponential generating function has a finite radius of convergence at the origin). This is a common phenomenon when trying to compute some series through the Euler-McLaurin summation formula. On the other hand, by stopping at $M=6$ (before Bernoulli numbers take off) the formula $(B)$ leads to $\mathcal{S}\approx 0.278092$, which is accurate.

Jack D'Aurizio
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