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.
 A: Here's another method by the Riemann integral, but not by definition:
$$
\sum_{n=1}^{\infty }(-1)^{n+1}\frac{1}{n}=
\lim_{m\to\infty}\sum_{n=1}^{m}(-1)^{n+1}\frac{1}{n}=
$$
$$
\lim_{m\to\infty}\int_0^1(1-x+\ldots+(-1)^{m-1}x^{m-1})\,dx=
\lim_{m\to\infty}\int_0^1\frac{1-(-x)^m}{1+x}\,dx=
\int_0^1\frac{dx}{1+x}=\ln2.
$$
A: 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
$$\sum_{n=1}(-1)^{n+1}\frac{1}{n}=\ln2.$$
A: 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
$$
\int_1^2\frac1x\,\mathrm{d}x=\log(2)\tag{2}
$$
A: As you asked for other methods of solving this problem, here is one using Taylor series.
Now that you know you are looking for the RHS of $\ln(2)$, expand $f(x) = \ln(x)$ into Taylor series around $x=1$:
$f^{(n)}(x) = (-1)^{n+1} x^{-n} (n-1)!,$ for $n>0, x \neq 0$
so choosing to evaluate at $x=1$, using $f(1) = \ln(1) = 0$, we get
$f(x) = f(1) + \sum_{n=1}^\infty \frac{f^{(n)}(1)(x-1)^n}{n!}
      = \sum_{n=1}^\infty (-1)^{n+1} \frac{(x-1)^n}{n}$
and using $x=2$,
$\ln 2 = \sum_{n=1}^\infty (-1)^{n+1}/n$
as desired.
A: However one should note that,
$$\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}$$
is not absolutely convergent as
$$\sum_{n=1}^\infty |(-1)^{n+1}\frac{1}{n}|=\sum_{n=1}^\infty \frac{1}{n}$$
is divergent. (Such a series is said to be conditionally convergent) 
Hence there is no fixed value for the series. (Refer to the Riemann Series Theorem)
$\ln 2$ is just a value that the series would take for the particular permutation of $S_{2n}$.
