Does the Taylor series of $\ln (1-x)$ converge uniformly on $[0, 1)$?

Question

We know that:

$$\ln (1-x) = -\sum_{n=1}^{\infty} \frac{x^n}{n}$$

Does the Taylor series converge uniformly on $[0, 1)$? I guess the answer is yes. What I have tried to do is that, after I showed pointwise convergence ,

$$\sum_{n=1}^{\infty} \frac{x^n}{n} - \sum_{n=1}^{N} \frac{x^n}{n} = \sum_{n=N+2}^{\infty} \frac{x^n}{n}$$

but I am unable to bound the last series. Any help to continue?

Edit: If it does not converge uniformly, how would one explain that $$\int_0^1 \sum_{n=1}^{\infty} \frac{x^n}{n}\, {\rm d}x= \sum_{n=1}^{\infty} \int_0^1 \frac{x^n}{n}\, {\rm d}x$$

Answer 1

For the first question:

Let $S$ any set, let $(f_n)_{n\in\mathbb N}$ a sequence of bounded functions (i.e. $\forall n,\, \sup_{x\in S}|f_n(x)|<+\infty$) converging uniformly on $S$ to a function $f$. Then $f$ is bounded as well.

So, the partial sums cannot converge uniformly on $[0,1)$ to $\ln(1-x)$, because $\lim_{x\to 1^-}|\ln(1-x)|=+\infty$

Answer 2

No, it does not converge uniformely on whole interval. But on every compact subset it does. $$ \forall \varepsilon > 0\; x \in <0, 1 - \varepsilon>: \ln(1 - x) \rightrightarrows T[\ln(1-x)](x) $$