Limit of another discontinuous function

This sum is the same as

HarmonicNumber $(2x)$ - HarmonicNumber $(x-1$ )

where

HarmonicNumber $(x)= \displaystyle \sum _{ k=1 }^{ x }{ \dfrac { 1 }{ k } }$ .

For large $x$ ,

HarmonicNumber $(x)=Log(x)+\gamma$

so the limit as $x \rightarrow \infty$ for this sum is $(Log(2x)+\gamma)-(Log(x-1)+\gamma)=Log(2)$

$\gamma$ is the Euler-Mascheroni constant.

First, since x is going to infinity, the floor functions can be basically ignored. Then, I made a not-so rigorous jump: I replaced the sum with an integral, since the value is very similar. At infinity, it turns out to be the same.

Then, I simply solved: $\lim_{x\to\infty}\int_x^{2x} \frac{1}{k}\mathrm{d}k = \lim_{x\to\infty}(\mathrm{ln}(2x)-\mathrm{ln}(x)) = \mathrm{ln}(2)$

So, the answer is ln 2, or $\boxed{.693}$