Why is $\lim\limits_{x\to\infty}\frac{\sum_{i=1}^x(\sum_{j=1}^i\frac1j-\ln i-\gamma)}{\sum_{i=1}^x\frac1i}=\frac12$

Question

$$
\mbox{Why is}\quad\lim_{x\to\infty}
\frac{\sum_{i = 1}^{x}\left[\sum_{j = 1}^{i}1/j -\ln\left(i\right)-\gamma\right]}{\sum_{i = 1}^{x}1/i} =
\frac{1}{2}\ ?.$$

I learnt Euler's Constant $\gamma$ before, and I want to know the sum of $H_k-\ln k-\gamma$. As Wikipedia says, $H_k=\ln k+\gamma+\varepsilon_k$, where $\varepsilon_k\sim\frac1{2k}$. But I wonder how to prove it. By the following Mathematica program, I can check that this limit is $\frac12$. But I cannot prove it. Why is it not anything below or above $\frac12$?

Limit[Sum[Sum[1/j, {j, 1, i}] - Log[i] - EulerGamma, {i, 1, x}]/Sum[1/i, {i, 1, x}], x -> Infinity]

Answer 1

With the help of @ParesseuxNguyen, I try to solve it with Stolz-Cesaro's lemma:

$\quad\lim\limits_{x\to\infty}\frac{\sum_{i=1}^x(\sum_{j=1}^i\frac1j-\ln i-\gamma)}{\sum_{i=1}^x\frac1i}\\ =\liminf\limits_{i\to\infty}\frac{\sum_{j=1}^i\frac1j-\ln i-\gamma}{\frac1i}\\ =\limsup\limits_{i\to\infty}\frac{\sum_{j=1}^i\frac1j-\ln i-\gamma}{\frac1i}\\ =\lim \limits_{i\to\infty}\frac{\sum_{j=1}^i\frac1j-\ln i-\gamma}{\frac1i}\\ =\lim \limits_{i\to\infty}\frac i{2i}=\frac12$.

(Since $H_i-\ln i-\gamma\sim\frac1{2i}$, so the last equality is right.)