prove that $\ln(2) = \sum_{k=1}^\infty \dfrac{(-1)^{k+1}}{k}$ without using Abel's theorem I know that
$$\ln(1+x) = \sum_{k=1}^\infty \dfrac{(-1)^{k+1}}{k}x^k$$
for $|x| < 1$. But how can I show
$$\ln(2) = \sum_{k=1}^\infty \dfrac{(-1)^{k+1}}{k}$$
without using Abel's theorem? There's a proof using Abel's theorem here. I know that if I define $s_0 = 1, s_1 := 1-1^2/2= 1/2$ and for $k\ge 2, s_k = s_{k-1} + \frac{(-1)^k}{k+1}$, then one can show by induction that $s_1 \leq s_{3}\leq \cdots s_{2n-1}\leq s_{2n}\leq s_{2n-2}\leq \cdots s_2 \leq s_0$. Also, if I let
$$L = \sum_{k=1}^\infty \dfrac{(-1)^{k+1}}k,$$
then for all $n, s_{2n+1}\leq L\leq s_{2n}\tag{1}.$ Taking $n = 1,$ we see that $s_3 = 1-1/2 + 1/3 \leq L \leq 1/2.$ (1) implies
$$|L-s_{2n}| \leq \frac{1}{2n+1}$$
and similarly
$$|L-s_{2n+1}| \leq \frac{1}{2n},$$
so sequences $(s_{2n})$ and $(s_{2n+1})$ both converge to the same limit $L$. But how can I use this to show the series converges to the required limit?
 A: Just for fun, here is a totally different approach without using the Taylor series of the logarithm:
$$\sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k} = 
\sum_{k=1}^{2n} \frac{1}{k} - 2 \sum_{\ell=1}^n \frac{1}{2 \ell} =  \sum_{k=1}^{2n} \frac{1}{k} -  \sum_{\ell=1}^n \frac{1}{\ell} = \sum_{k=n+1}^{2n} \frac{1}{k}.$$
So you are trying to compute $\lim_{n \to \infty} \sum_{k=n+1}^{2n} \frac{1}{k}$. We can rewrite this as $\lim_{n \to \infty} \frac{1}{n} \sum_{k=n+1}^{2n} \frac{1}{k/n}$. This last limit is a Riemann sum for $\int_1^2 \tfrac{dx}{x}$, which is $\log(2)$.
A: You can group terms together as
$$\sum_{n=1}^{\infty} \left( \frac{x^{2n-1}}{2n-1} - \frac{x^{2n}}{2n} \right) = \sum_{n=1}^{\infty} \frac{x^{2n-1} (1-x)}{(2n)(2n-1)}.$$
Then $x^{2n-1}(1-x) \leq 1$ for $0 \leq x \leq 1$, so this sum converges uniformly by the Weierstrass $M$-test. In particular, the sum is convergent, so the value of the sum at $x=1$ is the same as the limit as $x \to 1^{-}$.
This is pretty common for alternating series, and more generally for series whose signs have a periodic pattern, that you can group terms together to make them uniformly and absolutely convergent.
A: You can prove convergence to $\ln(2)$ with the estimate
$$\left|\sum_{k=1}^n \frac{(-1)^{k+1}}{k} - \ln(2)\right| \leqslant \frac{1}{n+1}$$

This follows easily by integrating the partial sum of a geometric series over $[0,1]$.  Note that
$$\sum_{k=0}^{n-1} (-1)^kx^k= \frac{1 - (-x)^n}{1-(-x)} = \frac{1}{1+x} + (-1)^{n+1} \frac{x^n}{1+x}$$
Hence,
$$\tag{1}\int_0^1 \left(\sum_{k=0}^{n-1} (-1)^kx^k\right) \, dx = \int_0^1\frac{dx}{1+x}+ (-1)^{n+1}\int_0^1\frac{x^n}{1+x} \, dx \\= \ln(2) +(-1)^{n+1}\int_0^1\frac{x^n}{1+x} \, dx $$
We can switch the integral with the finite sum on the LHS of (1) to obtain
$$ \tag{2}\int_0^1\left(\sum_{k=0}^{n-1} (-1)^kx^k\right) \, dx = \sum_{k=0}^{n-1} (-1)^k\int_0^1x^k\, dx = \sum_{k=0}^{n-1}\frac{(-1)^k}{k+1}= \sum_{k=1}^{n}\frac{(-1)^{k+1}}{k}$$
From (1) and (2) we get
$$\sum_{k=1}^{n}\frac{(-1)^{k+1}}{k} = \ln(2) + (-1)^{n+1}\int_0^1\frac{x^n}{1+x} \, dx$$
Whence,
$$\left|\sum_{k=1}^n \frac{(-1)^{k+1}}{k} - \ln(2)\right| = \left|(-1)^{n+1}\int_0^1\frac{x^n}{1+x} \, dx \right|= \int_0^1\frac{x^n}{1+x} \, dx\leqslant \int_0^1x^n \, dx = \frac{1}{n+1} $$
A: Noting for $x>0$,
\begin{eqnarray}
&&\bigg|\ln(1+x)-\sum_{k=1}^n \dfrac{(-1)^{k+1}}{k}x^k\bigg|\\
&=&\bigg|\int_0^x\frac{1}{1+t}dt-\sum_{k=1}^n(-1)^{k-1}\int_0^xt^{k-1}dt\bigg|\\
&=&\bigg|\int_0^x\bigg(\frac{1}{1+t}-\frac{1-(-t)^{n+1}}{1+t}\bigg)dt\bigg|\\
&=&\bigg|\int_0^x\frac{(-t)^{n+1}}{1+t}dt\bigg|\\
&\le&\int_0^xt^{n+1}dt=\frac{x^{n+2}}{n+2}
\end{eqnarray}
and letting $x\to1^-$, one has
$$ \bigg|\ln(2)-\sum_{k=1}^n \dfrac{(-1)^{k+1}}{k}\bigg|\le\frac{1}{n+2}. $$
So
$$ \ln(2)=\sum_{k=1}^\infty \dfrac{(-1)^{k+1}}{k}. $$
