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So, I'm supposed to use product of infinite series methods to find the series for $(\ln(1+z))^2$. I'm given that the answer has the form $$z^2 \sum_{l=0}^{\infty} c_l z^l$$ and I'm given that the first few terms are $$z^2-\left(1+\frac{1}{2}\right)\frac{2z^3}{3}+\left(1+\frac{1}{2}+\frac{1}{3}\right)\frac{2z^4}{4}-...$$

One can easily see that $c_l$ should be $\sum_{n=0}^{l}(-1)^l\frac{2}{(n+1)(l+2)}$, but I'd like to find this by working forward from the series rather than backward from its terms.

I start by squaring the series for $\ln(1+z)$: $$\left(\sum_{n=0}^{\infty}(-1)^n\left(\frac{z^{n+1}}{n+1}\right)\right) \left(\sum_{m=0}^{\infty}(-1)^m\left(\frac{z^{m+1}}{m+1}\right)\right)=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}(-1)^{n+m}\frac{z^{n+m+2}}{(n+1)(m+1)}\\=\sum_{l=0}^{\infty}\sum_{n=0}^{l}(-1)^{l}\frac{z^{l+2}}{(n+1)(l-n+1)}$$ where $l=n+m$.

This incorrectly suggests $c_l=\sum_{n=0}^l \frac{(-1)^l}{(n+1)(l-n+1)}$.

If anyone could spot my error, I would be very grateful.

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  • $\begingroup$ There is no error, you just need a small extra trick to convert your formula into the given one. $\endgroup$ – Jack D'Aurizio Jul 31 '15 at 21:15
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It is more practical to find the Taylor series of $\log^2(1-z)$, then replace $z$ with $-z$. Since:

$$ \log(1-z)=\sum_{n\geq 1}\frac{z^n}{n} $$ we have: $$ \log^2(1-z) = \sum_{n\geq 2}\left(\sum_{k=1}^{n-1}\frac{1}{k(n-k)}\right)z^n $$ but: $$ \frac{1}{k(n-k)}=\frac{1}{n}\left(\frac{1}{k}+\frac{1}{n-k}\right)$$ hence: $$ \log^2(1-z) = \sum_{n\geq 2}\frac{2 H_{n-1}}{n} z^n $$ where $H_m=\sum_{j=1}^{m}\frac{1}{j}$. That leads to:

$$ \log^2(1+z) = \sum_{n\geq 1}\frac{2(-1)^n H_{n-1}}{n} z^n.$$

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