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.

You must multiply the series
$$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+O\left(x^6\right)$$
$$\frac 1{1+x}=1-x+x^2-x^3+x^4-x^5+O\left(x^6\right)$$
$$\frac{\log(1+x)}{1+x}=\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+O\left(x^6\right) \right) \times$$ $$\left(1-x+x^2-x^3+x^4-x^5+O\left(x^6\right) \right) $$
$$\frac{\log(1+x)}{1+x}=x-\frac{3 x^2}{2}+\frac{11 x^3}{6}-\frac{25 x^4}{12}+\frac{137
   x^5}{60}+O\left(x^6\right)$$
In fact, almost as you wrote in the post
$$\frac{\log(1+x)}{1+x}=\sum_{n=1}^\infty (-1)^{n+1} H_n \,x^n$$
