Is $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots=\ln 2$ true? This might be an extremely stupid question but I stuck on two different definition of series expansion of $\ln(1+x)$. In  this Resonance article, the author assert that -
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots=\ln 2$$
based on the series expansion $\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\cdots  $ for $-1<x\leq1$. 
While the Wikipedia article says $\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\cdots  $ for $-1<x<1$.
Now these two definition looks confusing to me, as one definition includes the $1$ while other excludes it. If the first definition is true , then the sum of the series is correct, while other definition says it's not. 
Am I missing something obvious? Please can somebody explain this to me ? Thanks. 
 A: It is true and the Wikipedia article should have $\le 1$ at the end.  Expansions like this have a radius of convergence, in this case $1$.  For $x$ less than the radius the series converges absolutely.  For $x$ outside the series diverges.  When $x$ is equal to the radius it takes closer analysis.  Here the series is conditionally convergent as we can show
$$\sum _{i=1}^\infty \frac{(-1)^{i+1}}i=\sum_{i=1}^\infty \frac 1{2i-1}-\frac 1{2i}\\
=\sum_{i=1}^\infty\frac 1{2i(2i-1)}\\
\lt \sum_{i=1}^\infty \frac1{(2i-1)^2}$$
which we know converges.
A: Adding another way of proving the result. This requires less technology (without Abel's theorem). Instead I use a trick specific to this power series (well, it applies to some others as well).
Consider the series
$$
x-\frac {x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots
$$
in the interval $x\in(0,1]$. It is an alternating series where the sequence formed by the absolute values of the terms, $(x^n/n)$, A) decreases, B) converges to zero. Therefore, by Leibniz criterion on alternating series:


*

*The series converges to a limit $S(x)$ for all $x\in(0,1]$.

*The cut-off error $$R_n(x)=|S(x)-\sum_{k=1}^{n-1}\frac{(-1)^{k+1}x^k}k|<\frac{x^n}n\le\frac1n.$$


Item 2 shows that the series converges uniformly in the interval $x\in(0,1]$. Therefore we can conclude that $S(x)$ is continuous in that interval.
When $x\in(0,1)$ the series $S(x)$ is also obtained by integrating a uniformly converging geometric series term-wise. Therefore we know that $S(x)=\ln(1+x)$ whenever $x\in(-1,1)$.
The finishing touch: We know that both $S(x)$ and $\ln(1+x)$ are continuous in $(0,1]$. We also know that $S(x)=\ln(1+x)$ for all $x\in(0,1)$. By continuity
$$\ln(2)=\lim_{x\to1-}\ln(1+x)=\lim_{x\to1-}S(x)=S(1).$$
A: That the series expansion of $\ln(1+x)$ holds for $x=1$ is a consequence of Abel's Limit theorem
