Finding the limit as x approaches 1 of $\sum \frac {(-1)^{n+1}}{n} x^n$ I want to find $\displaystyle \lim_{x \rightarrow 1^-} \sum_{n \in \mathbb N} \frac {(-1)^{n+1}}{n} x^n$
What I'm guessing I'm supposed to do is use a certain property that states that if $\displaystyle \sum_{n \in \mathbb N} \frac {(-1)^{n+1}}{n} x^n$ converges for some $x$ in $[0,1]$ and the series of the derivatives $\displaystyle \sum_{n \in \mathbb N} (-1)^{n+1} x^{n-1}$ converges uniformly in $[0,1]$, then the original series converges uniformly and thus I can exchange the limit on $x$ and the limit on $n$, giving
$$\displaystyle \lim_{x \rightarrow 1^-} \lim_{N \rightarrow \infty} \sum_{n = 1}^N \frac {(-1)^{n+1}}{n} x^n = \lim_{N \rightarrow \infty} \lim_{x \rightarrow 1^-}  \sum_{n = 1}^N \frac {(-1)^{n+1}}{n} x^n$$
Which would result in the alternate Harmonic series (multiplied by $-1$).
The problem, however, is that while it is true that the derivatives converge uniformly on $[0,1)$, they don't on $[0,1]$ since when x = 1 the partial sums for the derivatives are $1,0,1,0 \dots$
However, the statement that the derivatives need to converge uniformly on $[0,1]$ is what's making my head spin. The original functions are defined on $[0,1]$ so shouldn't the derivatives be defined on $(0,1)$ ?. If not, what does it mean for the derivatives to converge uniformly on $[0,1]$ ?
 A: Let $f(x)= \sum\limits_{n=1}^{\infty} {(-1)^{n-1} x^{n-1}} $. This is geomertic seris and the sum is $\frac 1 {1+x}$. Now $\int_0^{x} f(t) dt=\sum\limits_{n=1}^{\infty} \frac {(-1)^{n-1} x^{n}} n$. Since $(-1)^{n-1}=(-1)^{n+1}$ the given sum is $\int_0^{x} f(t) dt=\ln (1+t)|_0^{x}=\ln (1+x)$. As $x \to 1$ this tends to $\ln 2$.
A: Here is one way forward.  We will write the series in terms of the even and odd parts of its summand.  Proceeding, we find that
$$\begin{align}
\sum_{n=1}^\infty \frac{(-1)^{n-1}x^n}{n}&=\sum_{n=1}^\infty \left(\frac{x^{2n-1}}{2n-1}-\frac{x^{2n}}{2n}\right)\\\\
&=\sum_{n=1}^\infty  x^{2n-1}\left(\frac{2n(1-x)+1}{2n(2n-1)}\right)\\\\
&=\sum_{n=1}^\infty \frac{(1-x)x^{2n-1}}{2n-1}+\sum_{n=1}^\infty \frac{x^{2n-1}}{2n(2n-1)}\tag1
\end{align}$$
Now, the second series on the right-hand side of $(1)$ is easily seen to be uniformly convergent for $0\le x\le 1$.
To show that the first series on the right-hand side of $(1)$ converges uniformly on $[0,1]$ we have for any $\varepsilon>0$ (and $N\ge1$)
$$\begin{align}
\sum_{n=N+1}^\infty \frac{(1-x)x^{2n-1}}{2n-1}&\le \sum_{n=N+1}^\infty \frac{(1-1/2n)^{2n-1}}{2n(2n-1)}\\\\
&\le \sum_{n=N+1}^\infty \frac{1}{4n(2n-1)}\\\\
&<\frac18\sum_{n=N+1}^\infty \left(\frac1{n-1}-\frac1n\right)\\\\
&=\frac1{8N}\\\\
&<\varepsilon
\end{align}$$
whenever $N>\frac1{8\varepsilon}$.   Hence, we can assure that
$$\begin{align}
\lim_{x\to 1^-}\sum_{n=1}^\infty \frac{(-1)^{n-1}x^n}{n}&=\sum_{n=1}^\infty \frac{1}{2n(2n-1)}\\\\
&=\sum_{n=1}^\infty \left(\frac{1}{2n-1}-\frac1{2n}\right)\\\\
&=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\\\\
&=\log(2)
\end{align}$$
as was to be shown!
