Prove the de la Vallée-Poussin's formula

With your current approach, you would have to split along the integers where $leftlceil frac{n}{k} rightrceil$ is constant, which is unnecessarily complex.

Instead, using Riemann summation, the limit $$lim_{n to infty} frac{1}{n} sum_{k=0}^n left( leftlceil frac{n}{k} rightrceil - frac{n}{k} right)$$

is $$int_0^1 left( leftlceil frac{1}{x} rightrceil - frac{1}{x} right) mathrm{d}x$$

More generally, $lim_{n to infty} frac{1}{n} sum_{k=0}^n fleft( frac{k}{n} right) = int_0^1 f(x) dx$. From here, split it into sections where $leftlceil frac{1}{x} rightrceil = n$. Equivalently, this means that $$n-1 < frac{1}{x} le n to frac{1}{n} le x < frac{1}{n-1}$$ The integral would then be $$sum_{n=2}^{infty} int_{frac{1}{n}}^{frac{1}{n-1}}left( n - frac{1}{x} right) mathrm{d}x$$

Evaluating the integral makes it $$sum_{n=2}^{infty}left(nleft(frac{1}{n-1}-frac{1}{n}right)-lnleft(frac{1}{n-1}right)+lnleft(frac{1}{n}right)right)$$

After simplifying a bit, it is $$sum_{n=2}^{infty}left(frac{1}{n-1}+lnleft(n-1right)-lnleft(nright)right) = sum_{n=1}^{infty}left(frac{1}{n}+lnleft(nright)-lnleft(n+1right)right)$$

Let the upper bound on the sum be $x$. Then this is $$lim_{x to infty} sum_{n=1}^{x}left(frac{1}{n}+lnleft(nright)-lnleft(n+1right)right)$$

The first term is $H_x$ and the rest telescopes to $ln(1) - ln(x+1) = -ln(x+1)$. Altogether, the limit becomes $$lim_{x to infty} left( H_x - ln(x+1) right)$$

You can then add $lim_{x to infty} left( ln(x+1)-ln(x) right) = 0$ to get $$lim_{x to infty} left( H_x - ln(x+1) right) + lim_{x to infty} left( ln(x+1)-ln(x) right) = lim_{x to infty} left( H_x-ln(x) right)$$

which is exactly the original definition of the Euler-Mascheroni constant.