There is a mistake in the following calculations. However, I can't find it so I'd like to ask for help \begin{align} S&=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{2n-1}-\frac{1}{2n}\\ &=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\dots+\frac{1}{2n-1}-\left(\frac{1}{2}+\frac{1}{4}+\dots+\frac{1}{2n}\right)\\ &=\sum_{k=0}^{n-1}\frac{1}{2k+1}-\frac{1}{2}\sum_{k=0}^{n-1}\frac{1}{k+1}=\sum_{k=0}^{n-1}\int_0^1 x^{2k} dx-\frac{1}{2}\sum_{k=0}^{n-1}\int_0^1 x^k dx\\ &=\int_0^1 \sum_{k=0}^{n-1} x^{2k} dx-\frac{1}{2}\int_0^1 \sum_{k=0}^{n-1} x^k dx=\int_0^1 \frac{1-x^{2n}}{1-x}dx -\frac{1}{2}\int_0^1 \frac{1-x^n}{1-x}dx\\ &=H_{2n}-\frac{H_n}{2} \end{align}
The result should've been
$$H_{2n}-H_{n}$$
By \begin{align} &1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\dots+\frac{1}{2n-1}-\left(\frac{1}{2}+\frac{1}{4}+\dots+\frac{1}{2n}\right)\\ &=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{2n}-\left(1+\frac{1}{2}+\dots+\frac{1}{n}\right)\\ &=H_{2n}-H_{n} \end{align}
$1 \over 1-x$ should be $1 \over 1-x^2$ on the fourth line.
$$ \sum_{k=0}^{n-1} x^{2k} = {1-x^{2n}\over 1-x^2} $$
Disclaimer: I didn't follow through with all the calculations to check that fixing this arrives at the correct answer.