In what sense does the Maclaurin series for $\ln(x+1)$ converge on $[-1,1]$?

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This is regarding Modes of Convergence. The question is as follows.

Given $f(x) = \ln(x+1)=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}$. Let $f_N(x)=\sum_{n=1}^{N}(-1)^{n-1}\frac{x^n}{n}$ Determine whether $f_N$ converges pointwise, uniformly, or in the $L^2$ sense to $f$ on $-1\leq x\leq1$.

I am having a hard time determining whether or not $f_N(x)$ does converge pointwise, uniformly, or in the $L^2$ sense to $f(x)$

I know $|f(x)-f_N(x)|\rightarrow 0$ as $N\rightarrow \infty$ for every $x$ in $-1<x<1$; however, I am not sure whether it converges pointwise or uniformly on the closed interval $[-1,1]$. The end point $-1$ is what causes the complications here. I want to say on the close interval, it does not converge uniformly because for all $\epsilon>0$, there does not exist a $N$ such that for all $n \geq N$, $|f(x)-f_N(x)|\leq \epsilon$ for all $x \in [-1,1].$

Any suggestions? Thanks in advance for your time.