I need to prove the following: $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+(-1)^{n+1}\frac{1}{n}+\cdots=\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}=\ln(2)$$

**Method 1:)**

The series $\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}$ is an alternating series, thus it is convergent, say to $l$. Therefore, both $s_{2n}$ and $s_n$ are convergent to the same limit $l$.

$$ \begin{align} s_{2n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots-\frac{1}{2n} & =\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{2n}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2n}\right) \\[10pt] & =\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} \end{align} $$

It is an easy exercise to prove that: $$\lim_{n \to \infty }s_{2n}=\lim_{n \to \infty }s_n =\lim_{n \to \infty }\left [ \frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} \right ]=\ln(2)$$ which implies that the given alternating series converges to $l=\ln 2$

However, I am interested to see a proof of this problem using the definition of the Riemann Integral as a sum of infinitely many rectangles of widths tending to zero. I tried to come up with a proof for this, but I couldn't. Can anyone share please?

Also, I am interested to see other methods of solving this problem (other than my method and the Riemann method). If anyone of you is aware of any other methods, please share :)