Finding the power series of $\frac{x}{1+\ln(1+x)}$ I've been stuck on trying to find the power series representation (order 3) for $\frac{x}{1+\ln(1+x)}$.
So far I know that there is the power series $\frac{1}{1-x}$ = $1 + x^2 + x^3 + x^4 + ...$
And I believe I can convert my original function to be $\frac{x}{1-(-\ln(1+x))}$ and then plug $\ln(1+x)$ wherever the $x$ is, yet I don't know how to simplify from there on. I checked my work on wolfram and found the answer should be $x-x^2+\frac{3}{2}x^3$, but I have no idea how to get there after plugging in $\ln(1+x)$, or if my original approach is even correct.
 A: Ignore the $x$ on top for the time being and first find the Maclaurin series of $\frac1{1+\log(1+x)}$. Let $y=\log(1+x)$:
$$\frac1{1+y}=1-y+y^2+O(y^3)$$
We only need to expand to $y^2$ (and to $x^2$ in $y$) since the $x$ we took out will eventually push the $y^3$ and higher terms into the $O$. For the same reason we only need the leading $x$-term at $y^2$:
$$\frac1{1+\log(1+x)}=1-(x-x^2/2)+x^2+O(x^3)=1-x+\frac32x^2+O(x^3)$$
$$\frac x{1+\log(1+x)}=x-x^2+\frac32x^3+O(x^4)$$
A: A bit late answer but I thought, here, it is worth mentioning the standard method shown below.
You may first calculate the first three terms of the power series of $\frac 1{1+\ln (1+x)}$ directly using the standard method of comparison of coefficients.
You have
$$1+\ln (1+x) = \sum_{n=0}^{\infty}a_n x^n \text{ with } a_0 = 1, a_n = \frac{(-1)^{n-1}}{n} \text{ for } n\geq 1$$
With $\frac 1{1+\ln (1+x)} = \sum_{n=0}^{\infty}b_n x^n$ the standard approach for the reciprocal is
$$\left(\sum_{n=0}^{\infty}a_n x^n\right)\left(\sum_{n=0}^{\infty}b_n x^n\right) = \sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}a_kb_{n-k}\right)x^n = 1$$
It follows:
$$a_0b_0 = 1 \Rightarrow b_0 = 1$$
$$a_1b_0 + a_0b_1 = 1 + b_1 = 0 \Rightarrow b_1 = -1$$
$$a_2b_0 +a_1b_1 +a_0b_2 = -\frac 12 -1 + b_2 = 0\Rightarrow b_2 = \frac 32$$
Hence, the first three terms in question are
$$\frac x{1+\ln (1+x)} = x -x^2 +\frac 32 x^3 + ...$$
