Does the function $H_{\lfloor x \rfloor} - \log(x)$ converge?

Question

$H_n$ here being the n-th harmonic number.

I know that the limes superior converges to the Euler-Mascheroni constant. However, it’s not clear to me whether the limes inferior of the function also equals this constant. Plotting the function in Mathematica first of all yields a very pretty picture, and secondly raises the question of whether the top and bottom curves seen converge to the same point:

Answer 1

Since$$ \varlimsup_{x → +∞} (H([x]) - \ln x) \leqslant \lim_{n → ∞} (H_n - \ln n) = γ,\\ \varliminf_{x → +∞} (H([x]) - \ln x) \geqslant \lim_{n → ∞} (H_n - \ln(n + 1)) = γ, $$ then$$ \lim_{x → +∞} (H([x]) - \ln x) = γ. $$

Answer 2

We have that for $n\leq x<n+1$ then $$H_{\lfloor x \rfloor} - \log(x)=H_n-\log(n+\{x\})=H_n-\log(n)-\log\left(1+\frac{\{x\}}{n}\right).$$ where $\{x\}:=x-\lfloor x \rfloor\in [0,1)$. Now note that, as $x$ goes to infinity, $n$ goes to infinity too and $$0\leq \log\left(1+\frac{\{x\}}{n}\right)\leq \log\left(1+\frac{1}{n}\right)\to 0$$ which implies that $H_{\lfloor x \rfloor}-\log(x)$ and $H_n-\log(n)$ has the same limit that is the Euler-Mascheroni constant $\gamma$.