the sum: $\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}=\ln(2)$ using Riemann Integral and other methods

Usually, to use the Riemann integral, alternating terms cause a problem. We want to have the finer partitions converge nicely, but an alternating series does not allow this. So, as far as I can see, we pretty much have to use the $\zeta$ trick you employ in your method: $$ \begin{align} &\frac11-\frac12+\frac13-\frac14+\dots+\frac1{2n-1}-\frac1{2n}\\ &=\left(\frac11+\frac12+\dots+\frac1{2n}\right)-2\left(\frac12+\frac14+\dots+\frac1{2n}\right)\\ &=\frac1{n+1}+\frac1{n+2}+\frac1{n+3}+\dots+\frac1{2n}\\ &=\sum_{k=n+1}^{2n}\frac{n}{k}\frac1n\tag{1} \end{align} $$ Then to use $(1)$ as a Riemann sum (with $x=k/n$ and $\mathrm{d}x=1/n$) for $$ \int_1^2\frac1x\,\mathrm{d}x=\log(2)\tag{2} $$

Here's another method by the Riemann integral, but not by definition: $$ \sum_{n=1}^{\infty }(-1)^{n+1}\frac{1}{n}= \lim_{m\to\infty}\sum_{n=1}^{m}(-1)^{n+1}\frac{1}{n}= $$ $$ \lim_{m\to\infty}\int_0^1(1-x+\ldots+(-1)^{m-1}x^{m-1})\,dx= \lim_{m\to\infty}\int_0^1\frac{1-(-x)^m}{1+x}\,dx= \int_0^1\frac{dx}{1+x}=\ln2. $$

You can use the following geometric series $$\sum_{n=0}^\infty x^n=\frac{1}{1-x},|x|<1.$$ Then integrate both sides to get $$\sum_{n=1}^\infty \frac{x^n}{n}=\int_0^x\frac{1}{1-t}dt=-\ln (1-x),|x|<1.$$ Since the LHS converges at $x=-1$, letting $x\to-1$ will give us $$\sum_{n=1}(-1)^{n+1}\frac{1}{n}=\ln2.$$