Why does $x/ln(x+1)$ = Something that includes factorials?

Question

So, I was doing some work when I saw I had to expand $\frac{x}{ln(1+x)}$

I know that expanding $ln(x+1) = \sum_{n=1}^ \infty [\frac{x^{2n-1}}{2n-1}-\frac{x^{2n}}{2n}] = \frac{x^1}{1}-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x^6}{6}+\frac{x^7}{7}-\frac{x^8}{8} ...$

But I got stuck when I had to divide $x$ by it.

I had to calculate the value of $\frac{x}{ \frac{x^1}{1}-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x^6}{6}+\frac{x^7}{7}-\frac{x^8}{8} ...}$

So I divided each term individually, and got :

$\frac{x}{x^1} - \frac{2x}{x^2}+\frac{3x}{x^3} - \frac{4x}{x^4}+\frac{5x}{x^5} - \frac{6x}{x^6}+\frac{7x}{x^7} - \frac{8x}{x^8}...$

When I checked on Wolfram Alpha, I got a totally different answer as shown in the link or below.

I definitely know I did something wrong, but I can't seem to identify it.

(Or in other words, what did I do wrong?)

P.S. I have a weird feeling that I did something wrong when dividing it from x.

Answer 1

Hint: Is $\frac 2{1+1} = \frac 21+\frac 21$?

Edit:

I'm not sure if there is a way to directly find whole Taylor expansion (I've glanced at the link JeanMarie provided in the comments and it seems it doesn't do it either, please correct me if I'm wrong). But there are ways to find Taylor polynomial up to any degree we desire.

Let $$\frac x{\ln(1+x)} = \sum_{n=0}^\infty a_nx^n,\quad \frac{\ln(1+x)}x = \sum_{n=0}^\infty b_nx^n.$$

Then we have $$\left(\sum_{n=0}^\infty a_nx^n\right)\left(\sum_{n=0}^\infty b_nx^n\right) = 1$$

and we can compare coefficients of the left and right hand sides to get

\begin{align} a_0b_0 &= 1\\ a_1b_0+a_0b_1 &= 0\\ a_2b_0+a_1b_1+a_0b_2 &= 0\\ &\ \ \vdots\\ a_nb_0+a_{n-1}b_1+\ldots a_1b_{n-1}+a_0b_n &= 0 \end{align}

Since we know all $b_n$, we can find $a_n$ recursively.

For example, if we want to find $a_n$ up to $n = 2$, we have:

\begin{align}a_0\cdot 1 = 1&\implies a_0 = 1\\ a_1\cdot 1+a_0\cdot\left(-\frac 12\right) = 0&\implies a_1 = \frac 12\\ a_2\cdot 1 + a_1\cdot \left(-\frac 12\right) + a_0\cdot \frac 13 = 0&\implies a_2 =-\frac 1{12}\end{align}

Another way to do the same thing is to calculate in quotient ring:

\begin{align}\frac x{x-x^2/2 + x^3/3+\ldots} + O(x^3) &= \frac{1}{1-x/2 + x^2/3} + O(x^3)\\ &=\left[\text{geometric series}\right]\\ &= \sum_{n=0}^\infty \left(\frac x2-\frac{x^2}3\right)^n + O(x^3)\\ &= 1 + \left(\frac x2-\frac{x^2}3\right) + \left(\frac x2-\frac{x^2}3\right)^2 + O(x^3)\\ &= 1 + \left(\frac x2-\frac{x^2}3\right) + \left(\frac {x^2}4- 2\cdot\frac{x}{2}\cdot\frac{x^2}3 + \frac{x^4}9\right) + O(x^3)\\ &= 1 +\frac x2 + \left(-\frac 1 3 + \frac 14\right)x^2+O(x^3)\\ &= 1 +\frac x2 -\frac {x^2}{12}+O(x^3)\end{align}