To prove: $\frac {\log (1+x)}{(1+x)} = \sum_{k=1}^∞ (-1)^{k+1}H_kx^k$

Question

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?

Answer 1

You need to use Cauchy product.

Let $F(x)=log(x+1)$ and $G(x)=(1+x)^{-1}$. Their product is the following. $$F(x)G(x)=\sum_{n=1}^\infty \sum_{k=1}^na_kb_{n-k}$$ Where $a_k=x^k(-1)^{k+1}k^{-1}$ and $b_{n-k}=(-x)^{n-k}$ $$\sum_{n=1}^\infty \sum_{k=1}^n(-1)^{k+1}x^kk^{-1}(-x)^{n-k}=\sum_{n=1}^\infty (-1)^{n+1}x^n\sum_{k=1}^nk^{-1}$$ As you can see the inner sum is equal to the harmonic number, so we can write: $$F(x)G(x)=\sum_{n=1}^\infty (-1)^{n+1}x^nH_{n}$$

Answer 2

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