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

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
$$\sum_{n=1}(-1)^{n+1}\frac{1}{n}=\ln2.$$
