Here is another method (not efficient) :

We use the fact that for all $x \in ]-1,1[$ , $\frac{1}{1+x}=\sum \limits_{n\ge 0} (-1)^n x^n$.

Then for all $x \in ]-1,1[$, we want to prove that : $\ln(1+x) =\sum \limits_{n\ge 0} (-1)^n \frac{x^{n+1}}{n+1}$.

Notice that for all $x \in[0,1[$, we have $\ln(1+x)= \int\limits_{0}^{x} \frac{1}{1+t}\mathrm{d}t$ and for all $x \in]-1,0]$, we have $\ln(1+x)= -\int\limits_{x}^{0} \frac{1}{1+t}\mathrm{d}t$. (Note that the function $t \mapsto \pm \frac{1}{1+t}$ is continuous on the compacts $[0,x]$ and $[x,0]$).

Now we can focus on the case where $x\in [0,1[$. The other case will be similar...

As the function $t \mapsto \frac{1}{1+t}\mathbb{1}_{[0,x]}(t)$ is positive and Lebesgue-measurable on $[0,1[$ we can write $\ln(1+x)= \int\limits_{0}^{1} \frac{1}{1+t}\mathbb{1}_{[0,x]}(t)\mathrm{d}t$.

Then $\ln(1+x)= \int\limits_{0}^{1} \sum \limits_{n\ge 0} (-1)^n t^n\mathbb{1}_{[0,x]}(t)\mathrm{d}t$ and we introduce for all $n\ge 0$ and for all $t\in [0,1[$ : $S_n(t,x)=\sum \limits_{k=0}^{n} (-1)^k t^k\mathbb{1}_{[0,x]}(t)$.

So $\ln(1+x)= \int\limits_{0}^{1} \sum \limits_{n\ge 0} (-1)^n t^n\mathbb{1}_{[0,x]}(t)\mathrm{d}t =\int\limits_{0}^{1} \lim\limits_{n\to+\infty} S_n(t,x)\mathrm{d}t$.

Then for all $n\ge 0$, the sequence of partial sums $S_n$ is Lebesgue-measurable on $[0,1[$ and for each $t\in[0,1[$ point-wise convergent to $S =\sum \limits_{n\ge 0} (-1)^n t^n\mathbb{1}_{[0,x]}(t)=\frac{1}{1+t}\mathbb{1}_{[0,x]}(t)$.

Moreover for all $n\ge 0$ and $t\in [0,1[$, we have $\vert S_n(t,x)\vert \le \sum \limits_{k= 0}^{n}t^k\mathbb{1}_{[0,x]}(t) \le \lim\limits_{n\to +\infty}\sum \limits_{k= 0}^{n}t^k\mathbb{1}_{[0,x]}(t)$. Because for all $k\ge 0$, the functions $t\mapsto t^{k}\mathbb{1}_{[0,x]}$ form a positive sequence of functions on $[0,1[$.
That's why $\vert S_n(t,x)\vert \le \frac{1}{1-t}\mathbb{1}_{[0,x]}(t)$.

The function $t\mapsto \frac{1}{1-t}\mathbb{1}_{[0,x]}(t)$ is positive and Lebesgue-measurable and even Lebesgue-integrable on $[0,1[$ because $\int \limits_{0}^{1} =\frac{1}{1-t}\mathbb{1}_{[0,x]}(t)\mathrm{dt} = \int \limits_{0}^{x}\frac{1}{1-t}\mathrm{d}t = \ln(1-x)<+\infty$

Then using the dominated convergence theorem we can write that :

$\ln(1+x)=\int\limits_{0}^{1} \lim\limits_{n\to+\infty} S_n(t,x)\mathrm{d}t = \lim\limits_{n\to+\infty} \int \limits_{0}^{1} S_n(t,x)\mathrm{d}t = \lim\limits_{n\to+\infty} \int\limits_{0}^{1}\sum \limits_{k=0}^{n} (-1)^k t^k\mathbb{1}_{[0,x]}(t)\mathrm{d}t$

$= \lim\limits_{n\to+\infty} \sum \limits_{k=0}^{n} (-1)^k \int\limits_{0}^{1} t^k\mathbb{1}_{[0,x]}(t)\mathrm{d}t = \lim\limits_{n\to+\infty} \sum \limits_{k=0}^{n} (-1)^k \int\limits_{0}^{x} t^k\mathrm{d}t = \lim\limits_{n\to+\infty} \sum \limits_{k=0}^{n} (-1)^k \frac{x^{k+1}}{k+1} = \sum \limits_{k=0}^{+\infty} (-1)^k \frac{x^{k+1}}{k+1}$.

With the same reasoning we can deduce the same result for $x\in ]-1,0]$.

Finally for all $x\in ]-1,1[$, we have $\ln(1+x) =\sum \limits_{n\ge 0} (-1)^n \frac{x^{n+1}}{n+1}$.

**NB** : **With properties on power series it takes four lines...**