Prove that if $|x| < 1$ then $\frac{\ln(1+x)}{1+x}=\sum_{n=1}^\infty (-1)^{n-1}s_nx^n,\,\,\,\,\,s_n\sum_{k=1}^n\frac{1}{k}$. By multiplying power series, show that if $|x| < 1$ then
$$\frac{\ln(1+x)}{1+x}=\sum_{n=1}^\infty (-1)^{n-1}s_nx^n,\,\,\,\,\,s_n\sum_{k=1}^n\frac{1}{k}.$$
So I know that:
$$\ln(1+x)=-\sum\limits_{n=1}^\infty\frac{(-1)^nx^n}{n},$$
when $|x|<1$, and:
$$\frac{1}{1+x}=\sum\limits_{n=0}^\infty (-1)^nx^n=1+\sum\limits_{n=1}^\infty (-1)^nx^n,$$
when $|x|<1$. So:
\begin{align*}
\frac{ln(1+x)}{1+x}={}&\left(-\sum\limits_{n=1}^\infty\frac{(-1)^nx^n}{n}\right)\left(1+\sum\limits_{n=1}^\infty (-1)^nx^n \right)={} \\
{}={}&-\sum\limits_{n=1}^\infty\frac{(-1)^nx^n}{n}+\left(-\sum\limits_{n=1}^\infty\frac{(-1)^nx^n}{n}\right)\left(\sum\limits_{n=1}^\infty (-1)^nx^n \right)={} \\
{}={}&\sum\limits_{n=1}^\infty\frac{-(-1)^nx^n}{n}+\left(\sum\limits_{n=1}^\infty\frac{-(-1)^nx^n}{n}\right)\left(\sum\limits_{n=1}^\infty (-1)^nx^n \right)={} \\
{}={}&\sum\limits_{n=1}^\infty\frac{-(-1)^nx^n}{n}+\sum\limits_{n=1}^\infty\left[\sum\limits_{k=1}^n\frac{-(-1)^{n-k}}{n-k}  (-1)^n\right]x^n={} \\
{}={}&\sum\limits_{n=1}^\infty\left[\frac{-(-1)^nx^n}{n}+ \sum\limits_{k=1}^n\frac{-(-1)^{n-k}}{n-k}  (-1)^n\right]x^n={} \\
{}={}&\sum\limits_{n=1}^\infty(-1)^{n-1}\left[\frac{-(-1)^nx^n}{n(-1)^{n-1}}+ \sum\limits_{k=1}^n\frac{-(-1)^{n-k}}{n-k}  \frac{(-1)^n}{(-1)^{n-1}}\right]x^n={} \\
{}={}&\sum\limits_{n=1}^\infty(-1)^{n-1}\left[\frac{1}{n}+ \sum\limits_{k=1}^n\frac{(-1)^{n-k}}{n-k} \right]x^n.
\end{align*}
Now I am stuck and not sure how to continue
 A: The following is not an answer but cannot be a comment either.
Let me take from your attempt:
\begin{align*}
\frac{\ln(1+x)}{1+x}={}&\left(-\sum\limits_{n=1}^\infty\frac{(-1)^nx^n}{n}\right)\left(1+\sum\limits_{n=1}^\infty (-1)^nx^n \right)={} \\
{}={}&-\sum\limits_{n=1}^\infty\frac{(-1)^nx^n}{n}+\left(-\sum\limits_{n=1}^\infty\frac{(-1)^nx^n}{n}\right)\left(\sum\limits_{n=1}^\infty (-1)^nx^n \right)={} \\
{}={}&\sum\limits_{n=1}^\infty\frac{-(-1)^nx^n}{n}+\left(\sum\limits_{n=1}^\infty\frac{-(-1)^nx^n}{n}\right)\left(\sum\limits_{n=1}^\infty (-1)^nx^n \right)={} \\
{}={}&\sum\limits_{n=1}^\infty\frac{-(-1)^nx^n}{n}+\sum\limits_{n=1}^\infty\left[\sum\limits_{k=1}^n\frac{-(-1)^kx^k}{k}(-1)^{n-k}x^{n-k}\right]={} \\
{}={}&\sum\limits_{n=1}^\infty\frac{-(-1)^nx^n}{n}+\sum\limits_{n=1}^\infty\left[\sum\limits_{k=1}^n\frac{-1}{k}(-1)^{n}x^{n}\right]={} \\
{}={}&\sum\limits_{n=1}^\infty\left[\frac{-1}{n}+\sum\limits_{k=1}^n\frac{-1}{k}  \right](-1)^nx^n={} \\
{}={}&\sum\limits_{n=1}^\infty(-1)^{n-1}x^n\left[\frac{1}{n}+\sum\limits_{k=1}^n\frac{1}{k}\right]x^n={}
\end{align*}
So the inside sum is indeed $s_n$, but we have a spurious $\frac1n$… If only the index of the inside sum stopped at $n-1$…
