How to derive Maclaurin series for ln(1+x) without calculus?

Question

How can we show $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$ for $-1 < x \leq 1$ without using calculus?

Answer 1

Without almost no calculus, perform the long division $$\frac{1}{1+x}=1-x+x^2-x^3+x^4+\cdots$$ and integrate both sides.

Answer 2

The power series for binomials $\displaystyle{(1+b)^x=\sum_{n=0}^{\infty}\frac{x(x-1)\dots(x-n+1)}{n!}b^n}$ and for exponent $\displaystyle{e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}}$ can be derived without Calculus.

Now, for a general $a=1+b$, you have $a^x=e^{(\ln a)x}=\sum_{n=0}^{\infty}\frac{(\ln a )^n (x)^n}{n!}$.

So $\ln a=\ln (1+b)$ is the coefficient of $x$ in the expantion of $a^x=(1+b)^x$, i.e. $$\displaystyle{\ln (1+b)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{b^n}{n}}.$$