Proof that $\displaystyle 1-\frac12+\frac13-\frac14+\dots=\ln2$.
I have read this answer but I don't understand the fifth line. \begin{align*} S_{2n}&=1-\frac12+\frac13-\frac14+\dots+\frac1{2n-1}-\frac1{2n} \\&=\left(1+\frac13+\dots+\frac1{2n-1}\right)-\left(\frac12+\frac14+\dots+\frac1{2n}\right) \\&=\left(1+\frac12+\frac13+\dots+\frac1{2n}\right)-2\left(\frac12+\frac14+\dots+\frac1{2n}\right) \\&=\left(1+\frac12+\frac13+\dots+\frac1{2n}\right)-\left(1+\frac12+\frac13+\dots+\frac1n\right) \\\color{gray}{[\text{ why? }]}\quad&=[\ln 2n+\gamma+o(1)]-[\ln n+\gamma+o(1)] \\&=\ln2+o(1) \end{align*} when $n\to\infty$, with $\gamma=\displaystyle\lim_{x\to\infty}\left(1+\frac12+\frac13+\dots+\frac1n-\ln n\right)$.
Otherway, $S_{2n+1}=S_n+\displaystyle\frac{1}{2n+1} \implies \lim_{n\to\infty}S_{2n+1}=\lim_{n\to\infty}S_{2n}=\ln2$
This proof uses the fact that $\lim_{n\to\infty} \left ( \left(\sum_{k=1}^n \frac{1}{k}\right) - \ln n\right )$ converges to a particular value, $\gamma\approx0.577...$ By using the definition of a limit, we can write:
$$ \gamma = \left(\sum_{k=1}^n \frac{1}{k}\right) - \ln n + \epsilon $$
Where $\epsilon$ can be made arbitrarily small by making $n$ larger and larger. (Values that behave this way can also be denoted by "$o(1)$")