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Using the following analytic continuation for the Alternating Harmonic Numbers ($\bar{H}_x=\sum_{i=1}^x\frac{(-1)^{i+1}}i$): $$\bar{H}_x=\ln2+\cos(\pi x)\left(\psi(x)-\psi\left(\frac x2\right)-\frac1x-\ln2\right)$$where $\psi$ is the digamma function, I was able to prove that the fractional part of the roots of this function approach $-\frac1\pi\arctan\left(\frac\pi{\ln2}\right)$ (as they approach $-\infty$). A reflection formula can be derived as well: $$\bar{H}_x-\bar{H}_{2-x}=\pi\cot(\pi x)+\left(\frac1{2-x}-\frac1{1-x}-\frac1x\right)\cos(\pi x)$$Using either the original definition, the analytic continuation, or Euler-Maclaurin definition (it's different from the one given above with its domain only being the positive real numbers) please post an answer with an identity/property for the Alternating Harmonic Numbers. It shouldn't be trivial (i.e. $\bar{H}_x=\bar{H_x}-1+1$ or $0\bar{H}_x=0$) though the proofs of the identities can involve them if needed.

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  • $\begingroup$ Have you considered using the definition $$\sum _{n=1}^x \frac{(-1)^{n+1}}{n}=\log (2)-(-1)^x \Phi(-1,1,x+1)$$ where $\Phi$ is the Lerch Transcendent (see WolframAlpha)? There are several identities for the Lerch Transcendent on the Wolfram Function Site. $\endgroup$ Commented May 27 at 19:57
  • $\begingroup$ @StevenClark That's essentially the analytic continuation I used to come up with the identities in OP. $\endgroup$ Commented May 27 at 20:06

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I'll make this answer to show how to find the asymptotic of the roots. We first observe that they must be non-positive. If we take the limit as $x\rightarrow-\infty$ in the reflection formula, we get that $\bar{H}_x-\bar{H}_{2-x}\sim\pi\cot(\pi x)$. Another important observation is that $\bar{H}_{2-x}\rightarrow\ln 2$ at the same time and so $\bar{H}_x\sim\pi\cot(\pi x)+\ln2$. Replacing $x$ with $x_n$, which is defined as the $n$th largest root, we get that $0\sim\pi\cot(\pi x_n)+\ln2$ or $\frac{\ln2}\pi\sim\cot(\pi x_n)$. Using IVT we can prove that there exists a root between the poles of $\bar{H}$, so we get $x_n\sim-n-\frac1{\pi}\arctan\left(\frac{\pi}{\ln2}\right)$.

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