# 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 $$[0,1]$$ ?

• Do you know measure theory and the dominated convergence theorem? Dec 8 '20 at 5:38
• If you want to establish uniform convergence, you can use this. Dec 8 '20 at 5:41
• You only need the derivatives to converge uniformly on $x \in [0,a]$ for each $0<a<1$, not $a=1$ (which would be false)... Dec 8 '20 at 5:44
• I know it would be false, but the text (reading this from Tao's analysis) says that the derivatives converge uniformly on the closed interval, which is kinda weird Dec 8 '20 at 6:00
• Sorry I havn't answered. I've been too focused on studying. I'll check your answer out when I can and tell you if I could follow it Dec 12 '20 at 0:04

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 $$\ln 2$$.

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!