Let $$f(x)=\lim_{n\to\infty}\text{Im}\left( -\frac{2\lfloor n\rfloor}{\pi}H_{-x}^{\left( -\frac1{2\lfloor n \rfloor} \right)} \right)$$ where $H$ is harmonic number and $x\in\mathbb{R}^+$. I do not know much about harmonic numbers, but I am wondering if this function is same as floor function, i.e. is it true that $f(x)=\lfloor x \rfloor$ for all $x\in\mathbb{R}^+$? I plotted the floor function and $f(x)$ for $n=1000$ and functions overlap. Is it just a coincidence or there is a proof that $f(x)=\lfloor x \rfloor$?
Edit
Also, I noticed that $f(x)$ is continuous on whole interval $(0,\infty)$ only if $n$ doesn't approach $\infty$. At $n\to\infty$ this function isn't continuous on integer values of $x$. Can it be proved?
OK, from the Mathematica code shown, I see that I had misinterpreted $H_{-x}^{(r)}$ as a power of $H_{-x}$. But in fact it is a two-argument generalized Harmonic number, denoted $H_{-x,r}$ on the Mathworld page.
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According to Maple, near $r=0$ we have $$ H_{k,r} = k - \big(\ln\Gamma(k+1)\big)r+O(r^2)\qquad\text{ as } r \to 0 $$ Therefore, for any $x > 0$ we have $$ \frac{-2n}{\pi}H_{-x,{-1/(2n)}} = \frac{-2n}{\pi}\left(-x+\ln\Gamma(-x+1)\frac{-1}{2n}+O\left(\frac{1}{n^2}\right)\right) $$ The $-x$ term is real, so we get $$ \lim_{n\to\infty}\mathrm{Im}\left(\frac{-2n}{\pi}H_{-x,{-1/(2n)}}\right) =\frac{-1}{\pi}\mathrm{Im}\big(\ln\Gamma(-x+1)\big) $$ So the question is whether this is $\lfloor x \rfloor$. It no longer has anything to do with harmonic numbers. It will follow from the functional equation for $\Gamma$.