To prove: $\frac {\log (1+x)}{(1+x)} = \sum_{k=1}^∞ (-1)^{k+1}H_kx^k$ For the open interval $(-1,1)$ we have:
$$\log (1+x) = x - \frac {1}{2}.x^2 + \frac {1}{3}.x^3 - ……$$
$$(1+x)^{-1} = 1 - x + x^2 - x^3 + ……$$
Multiplying the two series and collecting together the coefficients of like powers of $x$ we have:
$$\frac {\log (1+x)}{(1+x)} = x - (1+\frac {1}{2})x^2 + (1+\frac {1}{2} + \frac {1}{3})x^3 - …… =\sum_{k=1}^∞ (-1)^{k+1}H_kx^k$$
Here $H_n$ is the $n^{th}$ harmonic number. How can this result be proved analytically i.e. by the direct use of successive differentiation or by some other sounder method?

$$
\\\frac{\log{(1+x)}}{1+x}=\sum_{n=1}^{+\infty}{\frac{(-1)^{n+1}}{n}x^n}\sum_{m=0}^{+\infty}{(-1)^{m}x^m}=
\\=\sum_{n=1}^{+\infty}\sum_{m=0}^{+\infty}{\frac{(-1)^{n+m+1}}{n}x^{n+m}}=
\\=\sum_{n=1}^{+\infty}(-1)^{n+1}H_nx^n
\\k={n+m}=>n=1,\;2,\;3,\dots k=>
\\H_k=\sum_{n=1}^{k}{\frac{1}{n}}
$$
Q.E.D
