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Prove $\log (1+x) = \sum_{n=1}^\infty (-1)^{n+1}\frac{x^n}{n}$, $\forall x \in (-1,1]$

This question is supposed to be done using Taylor's theorem (not the one involving integrals):

For $ 0 < x \leq 1$ I used the Maclaurin's formula with the Lagrange's remainder and found the remainder to go to 0,

I am trying to deal with $ -1 < x < 0$ - It seems as though Lagrange's remainder isn't of much use so I'm trying to do this using Maclaurin's formula with Cauchy's remainder:

$R_n(x) =\dfrac{f^{(n+1)}(\xi)}{n!}(x-\xi)^nx$ for some $\xi \in (x,0)$ Using this and the fact that $f^{(n)}(x) = (-1)^{n-1}\dfrac{(n-1)!}{(1+x)^n}$ I get that

$$|R_n(x)| = \left | \dfrac{(x-\xi)^nx}{(1+\xi)^{n+1}} \right | = \left | \dfrac{(\xi-x)^nx}{(1+\xi)^{n+1}} \right | $$

The question gives a hint to consider the maximum of $\frac{\xi - x}{1+\xi}$ which I deduce to have a maximum of 1, so following on:

$$|R_n(x)| = \left | \dfrac{\xi - x}{1+\xi} \right |^n \cdot \dfrac{ |x| }{1+\xi} < \dfrac{|x|}{1+\xi}$$ but can't conclude that this goes to 0, any help, thank you.

from the comments, having the maximum as $|x|$ we get

$$|R_n(x)| = \left | \dfrac{\xi - x}{1+\xi} \right |^n \cdot \dfrac{ |x| }{1+\xi} < \dfrac{|x|^{n+1}}{1+\xi} < \dfrac{|x|^{n+1}}{1+ x} \to 0$$ as $n \to \infty$

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  • $\begingroup$ See this for one method. $\endgroup$
    – David Mitra
    Mar 12, 2014 at 12:54
  • $\begingroup$ How do you deduce the maximum is $1$? Since $x < \xi < 0$, the supremum is $\lvert x\rvert$. $\endgroup$
    – Daniel Fischer
    Mar 12, 2014 at 12:57
  • $\begingroup$ @DanielFischer How did you determine it's $|x|$? To determine the maximum I use $\xi \in (-1,0)$ and $ -1 < x < 0$ $\endgroup$
    – Warz
    Mar 12, 2014 at 13:12
  • $\begingroup$ You fix $x \in (-1,0)$. Then you use $\xi \in (x,0)$ and $$\frac{d}{d\xi} \frac{\xi-x}{1+\xi} = \frac{(1+\xi)\cdot 1 - 1\cdot(\xi-x)}{(1+\xi)^2} = \frac{1+x}{(1+\xi)^2} > 0$$ to conclude $0 < \frac{\xi-x}{1+\xi} < \frac{0-x}{1+0} = \lvert x\rvert$. $\endgroup$
    – Daniel Fischer
    Mar 12, 2014 at 13:18
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    $\begingroup$ Yes, it follows from there. Since the series doesn't converge uniformly on all of $(-1,0)$, we can't treat all these $x$ at once. Hence we must restrict the $x$ we consider. Instead of Fixing $x$, we could also have restricted it to an interval $[c,0)$ with $-1 < c < 0$, that would then show the uniform convergence on $[c,0)$. But looking at one $x$ at a time is conceptually simpler, so I started with that. $\endgroup$
    – Daniel Fischer
    Mar 12, 2014 at 13:38

1 Answer 1

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First, a power series $\sum c_n(x-a)^n$ has a convergence radius $R>0$, then there is integrable in $(R-a,R+a)$. The geometric series $\displaystyle \sum_{n=0}^{\infty}(-1)^nr^n$ has a convergence radius $R=1$, then we can be to integrate in $(-1,1)$, changing the simbols $\int$ and $\sum$: $\forall r\in (-1,1)$, $\displaystyle \frac{1}{1+r}=\frac{1}{1-(-r)}=\sum_{n=0}^{\infty}(-r)^n=\sum_{n=0}^{\infty}(-1)^nr^n$, then, $$\ln(1+x)=\int_{0}^x \frac{dr}{1+r}=\int_{0}^x\sum_{n=0}^{\infty}(-1)^nr^n dr=\sum_{n=0}^{\infty}(-1)^n\int_{0}^xr^ndr=$$ $$=\sum_{n=0}^{\infty}(-1)^n \frac{x^{n+1}}{n+1}=\sum_{N=1}^{\infty}(-1)^{N-1}\frac{x^N}{N},$$ i.e., $\displaystyle \ln(1+x)=\sum_{N=1}^{\infty}(-1)^{N-1}\frac{x^N}{N}$, for $x\in (-1,1)$.

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