taylor series of $\ln(1+x)$?

Solution 1:

You got the general expansion about $x=a$. Here we are intended to take $a=0$. That is, we are finding the Maclaurin series of $\ln(1+x)$. That will simplify your expression considerably. Note also that $\frac{(n-1)!}{n!}=\frac{1}{n}$.

The approach in the suggested solution also works. We note that $$\frac{1}{1+t}=1-t+t^2-t^3+\cdots\tag{1}$$ if $|t|\lt 1$ (infinite geometric series). Then we note that $$\ln(1+x)=\int_0^x \frac{1}{1+t}\,dt.$$ Then we integrate the right-hand side of (1) term by term. We get $$\ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots,$$ precisely the same thing as what one gets by putting $a=0$ in your expression.

Solution 2:

Note that $$\frac{1}{1+x}=\sum_{n \ge 0} (-1)^nx^n$$ Integrating both sides gives you \begin{align} \ln(1+x) &=\sum_{n \ge 0}\frac{(-1)^nx^{n+1}}{n+1}\\ &=x-\frac{x^2}{2}+\frac{x^3}{3}-... \end{align} Alternatively, \begin{align} &f^{(1)}(x)=(1+x)^{-1} &\implies \ f^{(1)}(0)=1\\ &f^{(2)}(x)=-(1+x)^{-2} &\implies f^{(2)}(0)=-1\\ &f^{(3)}(x)=2(1+x)^{-3} &\implies \ f^{(3)}(0)=2\\ &f^{(4)}(x)=-6(1+x)^{-4} &\implies \ f^{(4)}(0)=-6\\ \end{align} We deduce that \begin{align} f^{(n)}(0)=(-1)^{n-1}(n-1)! \end{align} Hence, \begin{align} \ln(1+x) &=\sum_{n \ge 1}\frac{f^{(n)}(0)}{n!}x^n\\ &=\sum_{n \ge 1}\frac{(-1)^{n-1}(n-1)!}{n!}x^n\\ &=\sum_{n \ge 1}\frac{(-1)^{n-1}}{n}x^n\\ &=\sum_{n \ge 0}\frac{(-1)^{n}}{n+1}x^{n+1}\\ &=x-\frac{x^2}{2}+\frac{x^3}{3}-... \end{align} Edit: To derive a closed for for the geometric series, let \begin{align} S&=1-x+x^2-x^3+...\\ xS&=x-x^2+x^3-x^4...\\ S+xS&=1\\ S&=\frac{1}{1+x}\\ \end{align} To prove in the other direction, use the binomial theorem or simply compute the series about $0$ manually.