Prove that $\lim\limits_{k\to\infty}\left(\frac1k\sum\limits_{n=1}^k\left\lfloor\frac kn\right\rfloor-\ln k\right)=2\gamma-1$

Note that

$$\tag{1}\frac1k\sum\limits_{n=1}^k\left\lfloor\frac kn\right\rfloor-\ln k = -\frac1k\sum\limits_{n=1}^k\left(\frac{k}{n}-\left\lfloor\frac kn\right\rfloor\right)-\left( \sum_{n=1}^k \frac{1}{n} - \ln k\right).$$

We have

$$\lim_{k \to\infty}\left( \sum_{n=1}^k \frac{1}{n} - \ln k\right) = \gamma,$$

and the first term on the RHS of (1) is a Riemann sum converging to

$$\lim_{k \to\infty}-\frac1k\sum\limits_{n=1}^k\left(\frac{k}{n}-\left\lfloor\frac kn\right\rfloor\right) = - \int_0^1 \left\{\frac{1}{x} \right\}\, dx, $$

where $\{ \cdot \}$ denotes fractional part.

It remains to show that the integral equals $\gamma - 1$.

This follows from

$$\begin{align}\int_0^1 \left\{\frac{1}{x} \right\}\, dx &= \int_1^\infty \frac{\{y\}}{y^2}\, dy \\ &= \sum_{k=1}^\infty \int_k^{k+1} \frac{y-k}{y^2} \, dy \\ &= \sum_{k=1}^\infty \left( \ln \frac{k+1}{k} - \frac{1}{k+1}\right) \\ &= \lim_{n \to \infty} \sum_{k=1}^n \left(\ln \frac{k+1}{k} - \frac{1}{k+1} \right) \\ &= \lim_{n \to \infty} \left(\ln (n+1) - \sum_{k=1}^n \frac{1}{k+1}\right) \\ &= 1 - \gamma \end{align}$$

Thus,

$$\lim_{k \to \infty} \left(\frac1k\sum\limits_{n=1}^k\left\lfloor\frac kn\right\rfloor-\ln k \right) = \gamma -1 + \gamma = 2 \gamma -1$$