Uniform convergence of power series of $\log(1+x)$ on $[0,1]$

Question

Define $$f(x): = \sum_{k = 1}^\infty \frac{(-1)^{k+1}x^k}{k},\;\;f_n(x) = \sum_{k = 1}^n \frac{(-1)^{k+1}x^k}{k}$$ $f(x)$ is a power series with radius of convergence 1, and $f(-1)$ diverges but $f(1)$ converges. I want to show $f$ is left continuous at $x = 1$.

My attempt:
I want to show $f_n\to f$ uniformly on $[0,1]$, and since all $f_n$ are continuous so will $f$ be on $[0,1]$. But to show uniform convergence, I refer to a lemma on Rudin's book pg 150 7.13:

$f_n\to f$ uniformly if:
(1) $f_n$ is a sequence of functions on compact interval $K$,
(2) $\;f_n\to f$ pointwise
(3) $\forall x\in K.\;f_n(x)\ge f_{n+1}(x).\;$

Unfortunately (3) does not hold for my example. But I am quite sure $f(x)$ is left-continuous at $x = 1$, how do I prove that? Thank you.

Answer 1

Since, for every $x$ in $[0,1]$, $f(x)$ is the sum of an alternating series (alternating signs + decreasing amplitudes), $$ |f(x)-f_n(x)|\leqslant\frac{x^{n+1}}{n+1}\leqslant\frac1{n+1}. $$ Thus, considering the supremum on $[0,1]$, $$ \|f-f_n\|_\infty\to0. $$