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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.)

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Write out the first couple of terms in the Taylor expansion of $\log(1+x)$ at $x=0$. Find $R_{0,k}$ and show that $R_{0,k} \to 0$ as $k \to \infty$. –  gt6989b May 6 '13 at 19:05
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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.

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