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 $[0,1]$ ?
Best Answer
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 $\ln 2$.