Taylor series of $[\log(1-z)]^2 $

Question

I'm having some trouble proving that the Taylor series about the origin of the function $[Log(1-z)]^2$ to be $$\sum_{n=1}^\infty \frac{2H_n}{n+1}z^{n+1}$$ where $$H_n = \sum_{j=1}^n \frac{1}{j}$$

So far, I've been trying to use the definition of Log(1-z) = log|1-z| + iArg(1-z). I've also noted that the Taylor expansion for log|1-z| = $\sum_{k=0}^\infty \frac{-z^{k+1}}{k+1}$ from integrating the geometric series $\sum_{k=0}^\infty z^k$ given |z|<1.

Because log|1-z| = $\sum_{k=0}^\infty \frac{-z^{k+1}}{k+1}$ converges for |z|<1, I've tried to use this for the Cauchy product of [Log(1-z)]^2

$$[Log(1-z)]^2 = Log(1-z) Log(1-z)$$ $$=\left(\sum_{i=0}^\infty \frac{-z^{i+1}}{i+1}\right)\left(\sum_{j=0}^\infty \frac{-z^{j+1}}{j+1}\right)$$ $$=\sum_{n=0}^\infty \sum_{k=0}^n\frac{-z^{k+1}}{k+1} \frac{-z^{n-k+1}}{n-k+1}$$ $$= \sum_{n=0}^\infty \sum_{k=0}^n\frac{z^{n+2}}{(k+1)(n-k+1)}$$

Really lost here, so any clues would be helpful! Thank you in advance.

Answer 1

The next step might be to use partial fractions:

$\dfrac1{k+1}+\dfrac1{n-k+1} =\dfrac{(n-k+1)+(k+1)}{(k+1)(n-k+1)} =\dfrac{n+2}{(k+1)(n-k+1)} $ so $\dfrac{1}{(k+1)(n-k+1)} =\dfrac1{n+2}\left(\dfrac1{k+1}+\dfrac1{n-k+1}\right) $.

Therefore

$\begin{array}\\ \sum \limits_{n=0}^\infty \sum \limits_{k=0}^n\dfrac{z^{n+2}}{(k+1)(n-k+1)} &=\sum \limits_{n=0}^\infty \sum \limits_{k=0}^nz^{n+2}\dfrac1{n+2}\left(\dfrac1{k+1}+\dfrac1{n-k+1}\right)\\ &=\sum \limits_{n=0}^\infty \dfrac{z^{n+2}}{n+2}\sum \limits_{k=0}^n\left(\dfrac1{k+1}+\dfrac1{n-k+1}\right)\\ &=\sum \limits_{n=0}^\infty \dfrac{z^{n+2}}{n+2}\left(\sum \limits_{k=1}^{n+1}\dfrac1{k}+\sum \limits_{k=0}^{n}\dfrac1{n-k+1}\right)\\ &=\sum \limits_{n=0}^\infty \dfrac{z^{n+2}}{n+2}\left(H_{n+1}+\sum \limits_{k=1}^{n+1}\dfrac1{k}\right)\\ &=\sum \limits_{n=0}^\infty \dfrac{z^{n+2}}{n+2}2H_{n+1}\\ &=\sum \limits_{n=1}^\infty \dfrac{z^{n+1}}{n+1}2H_{n}\\ \end{array} $

Answer 2

Outline: I would attack the problem somewhat differently. First multiply the power series for $\ln(1-t)$ by the series for $\frac{2}{1-t}$. The $2H_k$ will come up naturally as coefficients.

Then integrate term by term from $0$ to $z$. That will produce the right denominator.

Finally, calculate $\int_0^z\frac{2}{1-t}\ln(1-t)\,dt$.