The questions is as defined below.
Let $f(x)= \log(1+x)$. Show that the Taylor remainder $R_{0,k}(x)$, defined by $$R_{a,k}(x)= f(a+x) - P_{a,k}(x) = f(a+h) -\sum_{j=0}^{k} \frac {f^{j}(a)x^{j}}{j!},$$ tends to $0$ as $k \to \infty$ for $ -1<x \leq 1$. Conclude that $$\log(1+x)= \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n}$$ for $-1 < x \leq 1$.
Pretty much all I got here is "lol wut"? (I tried this question a few times, but I have no idea how to even approach it.)
One can write the Taylor remainder as an integral :
$$ f(a+x) -\sum_{j=0}^{k} \frac {f^{(j)}(a)x^{j}}{j!}= \int_{a}^{a+x} \frac{(a+x-t)^k}{k!} f^{(k+1)}(t)dt \tag{1} $$
Taking $f(x)={\sf log}(x)$ and $a=1$, this yields
$$ R_{0,k}(x)={\sf log}(x)-\Bigg(\sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n}\Bigg)= (-1)^{k} \int_{1}^{1+x} \left(\frac{1+x-t}{t}\right)^k dt \tag{2} $$
When $t$ is between $1$ and $1+x$, we have
$$ \bigg| \frac{1+x-t}{t} \bigg| \leq |x| \tag{3} $$
And hence
$$ |R_0(x)| \leq \int_{1}^{1+x} \left(\frac{|1+x-t|}{|t|}\right)^k dt \leq \int_{1}^{1+x} |x|^{k} dt =|x|^{k+1} \tag{4} $$
When $|x| \lt 1$, this tends to zero when $k\to \infty$ and we are done. The only case left is $x=1$. Then
$$ R_{0,k}(1)= \int_{1}^{2} \left(\frac{2-t}{t}\right)^k dt \tag{5} $$
Putting $t=\frac{2}{z+1}$, we see that
$$ R_{0,k}(1)= 2\int_{0}^{1} \frac{z^k}{(z+1)^2} dz \leq 2\int_{0}^{1} z^k dz =\frac{2}{k+1} \tag{6} $$
This still converges to zero, which concludes the proof.