Error Analysis and Modes of Convergences I have the following question regarding Modes of Convergence


Recall that the Taylor series of the function $f(x) = \ln(1+x)=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}$. Let $f_{N}(x)=\sum_{n=1}^{N}(-1)^{n-1}\frac{x^n}{n}$ be the partial sum. Determine whether $f_N$ converges pointwise, uniformly, or in the $L^2$ sense to $f$ on $-1\leq x\leq1$.


I am kinda sketchy on whether it converges pointwise - well in general actually - for $-1$ since $\ln(1+-1) = \ln(0)$ is undefined. If I disregard this, let's take a look at what we know. $\ln(x+1)$ = $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}$ converges on $|x|<1$ by the ratio test. The ends would be of concern, but $1$ is okay. Not so much $-1$. Anywho, this is what I have.
Pointwise Converges:
Consider $f(x) = \sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}$ and its $N$th partial sum $f_{N}(x)=\sum_{n=1}^{N}(-1)^{n-1}\frac{x^n}{n}$. The infinite series converges pointwise to $f(x)$ on $-1\leq x \leq 1$ if 
$
\begin{eqnarray} 
| \sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}-\sum_{n=1}^{N}(-1)^{n-1}\frac{x^n}{n}|\\
|(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+...)-(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+...)|\\
|(x-x)+(-\frac{x^2}{2}+\frac{x^2}{2})+(\frac{x^3}{3}-\frac{x^3}{3})+(-\frac{x^4}{4}+\frac{x^4}{4})+(\frac{x^5}{5}-\frac{x^5}{5})+...| \longrightarrow 0\\
\end{eqnarray}
$
as $N \rightarrow \infty$, for every $x\in[-1,1]$.
However, for some reasoning I am doubting it actually converges on the entire interval. I could just be over thinking it. The point is to show the error between $f(x)$ and $f_N(x)$ tends to zero as $N\rightarrow \infty$ for every $x$ on the interval and it seems to do just that.
Uniformly Convergence is similar; however, it focuses on uniform error between $f(x)$ and $S_N(x)$ tends to zero as $N\rightarrow \infty$. If want to conclude it also tends to zero. Same goes for $L^2$ sense.
Any suggestions? Thank you for your time and Thanks in advance for any feedback. 
 A: Here is a start. You already established pointwise convergence for $|x| < 1$.
When $x = -1$, you have a divergent harmonic series $\sum 1/n $.
When $x = 1$, you have a convergent alternating series $\sum (-1)^n/n.$ 
If $|x| \leq r <1$, then
$$\left|\frac{(-1)^{n} x^n}{n}\right|\leq \frac{r^n}{n} < r^n.$$
The geometric series $\sum r^n$ converges for $|r| < 1$:
$$\sum_{n=1}^{\infty}\left|\frac{(-1)^{n} x^n}{n}\right|<\sum_{n=1}^{\infty}r^n = \frac{r}{1-r}.$$
Therefore, the series converges absolutely and uniformly (by the Weierstrass M-test) on any compact interval $[-r,r]$ with $r < 1$.
Looking at it more closely, we have uniform convergence on $[-r,r]$ because the sequence of partial sums of $\sum r^n$ converges and constitutes a Cauchy sequence.
Thus, for any $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that if $m \geq n \geq N$, then for all $x \in [-r,r]$
$$|S(m) - S(n)|=\left|\sum_{k=n=1}^{m}\frac{(-1)^{n} x^n}{n}\right| \leq \sum_{k=n=1}^{m}\left|\frac{(-1)^{n} x^n}{n}\right|<\sum_{k=n+1}^{m}r^k < \epsilon.$$
Hence, the sequence of partial sums satisfies the Cauchy criterion for convergence uniformly.
A more subtle question is does the series converge uniformly on $(-1,1]$. This is not the same as local uniform convergence on every $[-r,r]$ with $r < 1$.
