Elementary proofs for Taylor expansion for natural logarithm

Question

For $|x|<1$, we have that

$$ \ln(1+x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \cdots $$

Is there any elementary(try not to use integration or differantiation) proof for the equality above?

Edit: I have changed my definition of elemantary.

Answer 1

Consider

$$ \frac{1}{1-x} = \sum_{n \geq 0 } x^n $$

Thus,

$$ \frac{1}{1+x} = \sum_{n \geq 0 } (-1)^nx^n $$

Integrating, we obtain

$$ \ln (1+x) + C = \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1} }{n+1} $$

With $x=0$ , $C$ better be zero and thus

$$ \ln (1+x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1} }{n+1} $$

Answer 2

Here is a sketch, which if extended will find any particular coefficient without calculus, on the assumption a polynomial expression exists

Let's start with the definitions:

• $\exp(x)= 1+\frac{1}{1!}x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\cdots$ and
• for positive $x$ we have $\exp(\ln(x))=x$

Suppose $\ln(1+x)=a_0+a_1x+a_2x^2+a_3x^3+\cdots$

Clearly $a_0=0$ since $\exp(0)=1$ and so $\ln(1+0)=0$

Then we need $1+x = 1 + \frac{1}{1!}(a_1x+a_2x^2+a_3x^3+\cdots)+\frac{1}{2!}(a_1x+a_2x^2+\cdots)^2+\frac{1}{3!}(a_1x+\cdots)^3+\cdots$

which by

• matching coefficients of $x$ will give $a_1=1$, and
• matching coefficients of $x^2$ will give $a_2+\frac{a_1^2}{2!}=0$ so $a_2=-\frac12$, and
• matching coefficients of $x^3$ will give $a_3+\frac{2a_1a_2}{2!}+\frac{a_1^3}{3!}=0$ so $a_3=+\frac13$, and
• matching coefficients of $x^4$ will give $a_4+\frac{2a_1a_3+a_2^2}{2!}+\frac{3a_1^2a_2}{3!}+\frac{a_1^4}{4!}=0$ so $a_4=-\frac14$, and
• we can do something similar for later coefficients