Let us consider the function:
$$ f(α,β) \equiv \sum_{n = 1}^{\infty}\left(-1\right)^{n - 1}\left[% {n^{2\alpha - 1} - 1 \over n^{\alpha}}\,\cos\left(\beta\ln\left(n\right)\right) \right] $$
My question is: Show that $f$ is harmonic $\quad\forall\ s = \alpha + \beta\,{\rm i}\quad$ with $\quad 0 < \alpha < 1$.
Yes, your analysis seems to be correct. $$ \begin{align} f(\alpha,\beta) &=\sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{n^{2\alpha-1}-1}{n^\alpha}\right)\cos(\beta\log(n))\\ &=\mathrm{Re}\left(\sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{n^{2\alpha-1}-1}{n^\alpha}\right)n^{i\beta}\right)\\ &=\mathrm{Re}\left(\sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac1nn^{\alpha+i\beta}-\frac1{n^{\alpha-i\beta}}\right)\right)\\[9pt] &=\mathrm{Re}\left(\eta(1-s)-\eta(\bar{s})\right) \end{align} $$ Analytic and conjugate analytic functions are harmonic as is their real part.
Since I plotted $f$ for a comment, I include it here. The red line is $\left(\frac12+iy,f\left(\frac12+iy\right)\right)$
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