Prove $\frac{\ln(1+x)}{1+x}=x-\left(1+\frac{1}{2}\right)x^2+\left(1+\frac{1}{2}+\frac{1}{3}\right)x^3-\ldots$ I have to prove that :
$$\frac{\ln(1+x)}{1+x}=x-\left(1+\frac{1}{2}\right)x^2+\left(1+\frac{1}{2}+\frac{1}{3}\right)x^3-\ldots$$
I already know that the Maclaurin series of $\ln(1+x)$ is $x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+\frac{1}{5}x^5+\ldots$ But how can I use it to prove the first claim? Should I just multiply $\frac{1}{1+x}$ to the Maclaurin series of $\ln(1+x)$? Still, I don't know how to do that.
I tried to use the derivatives of $\frac{\ln(1+x)}{1+x}$ then substitute that into the Maclaurin formula. But I don't think that's the right way to do this efficiently.

Multiply the two series $\ln(1+x) = x - \frac{1}{2} x^2 + \frac{1}{3} x^3 - \cdots$ and $\frac{1}{1+x} = 1 - x + x^2 - x^3 + \cdots$.
For instance, the coefficient of $x^2$ in the product is $(-\frac{1}{2} x^2) \cdot 1 + x \cdot (-x) = -(1 - \frac{1}{2}) x^2$.
