$\frac12 [log(1-x)]^2 = \frac12 x^2 + (1+ \frac12) \frac13 x^3+ (1+ \frac12 + \frac13) \frac14 x^4+ \cdots$
My attempt: I'm thinking of finding a series which is convergent in $-1<x<1$ and integrating term by term to obtain my required expression. So, I differentiated the given RHS to get the series
$x+ (1+ \frac12) x^2 + (1+ \frac12 + \frac13) x^3+ \cdots$
So, the $n^{th}$ term is $(1+ \frac12+ \cdots + \frac1n) x^n$
This is where I'm stuck. How do I prove it's convergent and how do I find it's sum? I tried Ratio Test, I couldn't find the limit. Please help!
To show convergence by the ratio test, noted that the coefficient of $x^n$ is $$ a_n = \frac{1}{n} H_{n-1} \leq \frac{1}{2} $$ for all $n \geq 1$. ($H_n$ is defined as $\sum_{k=1}^n(1/k)$.) So the ratio $|a_{n+1}/a_n| \leq x/2$ which is less than $\frac{1}{2}$ so the series converges when $|x| < 2$. (Although the interpretation in terms of a log is tricky when $x < -1$.)
To prove the coeffiecients are the ones given, start from $$ \log(1-x) = - \sum_1^\infty \frac{1}{n}x^n$$ which is easily shown as a Taylor series. Then, as long as the final series can be shown to converge, $$ [\log(1-x)]^2 = \sum_{n=1}^\infty \frac{1}{n}x^n \sum_{m=1}^\infty \frac{1}{m}x^m = \sum_{m,n > 0} \frac{1}{mn}x^{m+n} $$ Transform indices to use $r = m+n$: $$ [\log(1-x)]^2 = \sum_{r > n > 0} \frac{1}{r-n}\frac{1}{n}x^{r} = \sum_{r=2}^\infty \sum_{n=1}^{r-1} \frac{1}{r-n}\frac{1}{n} $$ Now we need to evaluate $\sum_{n=1}^{r-1} \frac{1}{r-n}\frac{1}{n}$ but instead we will evaluate $$ r\sum_{n=1}^{r-1} \frac{1}{r-n}\frac{1}{n} = \sum_{n=1}^{r-1}\left( \frac{r}{r-n}\right)\frac{1}{n} = \sum_{n=1}^{r-1}\frac{1}{n}+\sum_{n=1}^{r-1}\frac{n}{r-n}\frac{1}{n} = H_{r-1} + \sum_{n=1}^{r-1}\frac{1}{r-n} = 2H_n $$ So $$ \frac{1}{2}\sum_{n=1}^{r-1} \frac{1}{r-n}\frac{1}{n} = \frac{1}{r}H_{r-1}$$ and $$ \frac{1}{2} [\log(1-x)]^2 = \sum_{n=1}^\infty \frac{1}{r}H_{r-1}x^r $$ which is the expression wanted. These manipulations are valid only if the final expression converges, but we showed that at the beginning.