How to verify these series expansions? We know that $$ x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \cdots (-1)^n \frac{x^{n+1}}{n+1} + \cdots = \log (1+x) $$ whenever $-1<x<1$. What can we say if $x=1$? 
That is, does the series $$ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \cdots (-1)^n \frac{1}{n+1} + \cdots $$ converge to $\log 2$? If so, how to prove this fact rigorously? 
And what about the behavior of the series $$ x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \cdots (-1)^n \frac{x^{2n+1}}{2n+1} + \cdots, $$
which converges to $\arctan x$ for $-1<x<1$, at the points $x = \pm 1$? 
 A: Daniel's comment already has the solution: since
$$\sum_{n=1}^\infty \frac{(-1)^{n-1}}n$$
converges (Leibniz series) and the power series
$$\sum_{n=1}^\infty\frac{(-1)^{n-1}x^n}n=f(x)$$
converges for $\,|x|<1\,$ to a function $\,f(x)\;$ , then Abel's Theorem gives us
$$\lim_{x\to 1^-}f(x)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}n$$
from here, the sum of the above series indeed is 
$$\lim_{x\to 1^-}\log(1+x)=\log2\;$$ since, of course, $\,f(x)=\log(1+x)\;$
A: Note that for $x\lt1$
$$
\frac{x^{2k-1}}{2k-1}-\frac{x^{2k}}{2k}\gt0
$$
and
$$
\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{x^{2k-1}}{2k-1}-\frac{x^{2k}}{2k}\right)
=x^{2k-2}-x^{2k-1}\gt0
$$
Thus, as $x\to1^-$,
$$
f(x)=\sum_{k=1}^\infty\left(\frac{x^{2k-1}}{2k-1}-\frac{x^{2k}}{2k}\right)
$$
is a positive series, increasing termwise to
$$
f(1)=\sum_{k=1}^\infty\left(\frac1{2k-1}-\frac1{2k}\right)
$$
which converges. Thus, by dominated convergence,
$$
\lim_{x\to1^-}f(x)=f(1)
$$
The same argument works for
$$
g(x)=\sum_{k=1}^\infty\left(\frac{x^{4k-3}}{4k-3}-\frac{x^{4k-1}}{4k-1}\right)
$$
