Now $$e_n-e_0=\sum_{k=0}^{n-1}\left [ -\frac{1}{12(k+x)^2}+\mathcal{O}\left ( \frac{1}{(k+x)^3} \right ) \right ]; \tag{*}$$ therefore, $\lim_{n\to\infty}e_n-e_0=K_1(x)$ exists. Set $$e_n=K(x)+\frac{1}{12(n+x)}+\mathcal{O}\left ( \frac{1}{(n+x)^2} \right ),$$ where $K(x)=K_1(x)+e_0$. The term $(n+x)^{-1}$ comes from completing the sum in $(*)$ to infinity and approximating the added sum by an integral.
This is taken from a book on page 19. This is about estimating a specific value (integral). I do not understand the two last sentences. Could you please explain it for me?
They are using the Euler–Maclaurin formula which states that the approximation $$\sum_{i=n}^mf(i)\approx\int_n^mf(x)dx$$ holds (see here for further details) and combining it with the fact that $\sum_{i=N}^\infty f(i)$ is small for any convergent summation. So you can approximate the sum with it's "completed" infinite sum, and then approximate the infinite sum with it's integral.