Show that $\frac{1}{2}(0.5)+\frac{(1+\frac{1}{2})}{3}(0.5)^2+\frac{(1+\frac{1}{2}+\frac{1}{3})}{4}(0.5)^3+\cdots=(\log 2)^2$ Show that $$\frac{1}{2}(0.5)+\frac{(1+\frac{1}{2})}{3}(0.5)^2+\frac{(1+\frac{1}{2}+\frac{1}{3})}{4}(0.5)^3+\cdots=(\log 2)^2$$
I tried to prove it by using the Taylor series, But I've found it not useful as shown below

any help, thanks  
 A: Note that your series is $f(\frac{1}{2})$, where 
$$f(x) = \sum_{n=1}^\infty \frac{H_n}{n+1} x^n$$
($H_n$ being the hamonic series), which is a power series of radius of convergence $1$. Now, $x f(x)=\sum_{n=1}^\infty \frac{H_n}{n+1} x^{n+1}$ looks like (and is) the antiderivative of
$$F(x) = \sum_{n=1}^\infty H_n x^n$$
on $(-1,1)$, by properties of power series. So one can first try to find a closed form for $F$. Either because of a hunch ($H_n$ as coefficient looks like the coefficient of the Cauchy product between series of coefficients $1$ and $\frac{1}{n}$ respectively, which are known), or looking at the output of Mathematica (or both), we can guess what to do. (More details follow, with a full derivation.)

We will use the fact that, for $\lvert x\rvert < 1$,
$$\begin{align}
-\ln(1-x) &= \sum_{n=1}^\infty \frac{x^n}{n} \\
\frac{1}{1-x} &= \sum_{n=0}^\infty x^n
\end{align}$$
so that
$$\begin{align}
\frac{-\ln(1-x)}{1-x} &= \sum_{n=1}^\infty \left(\sum_{k=1}^n 1\cdot \frac{1}{k}\right) x^n = \sum_{n=1}^\infty H_n x^n
\end{align}$$
as wished (where, in the middle, we used the fact that $\left(\sum_{n=0}^\infty a_n x^n\right) \cdot \left(\sum_{n=0}^\infty b_n x^n \right) = \sum_{n=0}^\infty c_n x^n$ with $c_n = \sum_{k=0}^n a_k b_{n-k}$). Now, computing the antiderivative:
$$
F(x) - F(0) = F(x) = \int_0^x \frac{-\ln(1-t)}{1-t}dt = \frac{1}{2}\ln^2(1-x)
$$
(recognizing an integral of the form $\int_0^x u^\prime u = \frac{1}{2}[u^2]^x_0$, for $u(x) = -\ln(1-x)$). 
This implies that, for $\lvert x\rvert < 1$, 
$$xf(x) = \frac{1}{2}\ln^2(1-x)$$
so that, for $x\in(0,1)$,
$$f(x) = \frac{\ln^2(1-x)}{2x}.$$
Plugging in $x=\frac{1}{2}$ leads to $$
f\!\left(\frac{1}{2}\right) =  \ln^2 \frac{1}{2} = \ln^2 2.
$$
A: $$\begin{align}
\sum_{n=2}^{\infty} \frac{H_{n-1}}{n 2^{n-1}}=\sum_{n=2}^{\infty}\sum_{m=1}^{n-1} \frac1{n 2^{n-1} m}
\\=\sum_{m=1}^{\infty}\sum_{n=m+1}^{\infty}\frac1{n 2^{n-1} m}
\\=2\sum_{n,m=1}^{\infty}\frac1{m(n+m)2^n\,2^m}
\\=2\sum_{n,m=1}^{\infty}\frac1{n(n+m)2^n\,2^m}
\\=\sum_{n,m=1}^{\infty}\frac1{n 2^n}\frac1{m 2^m}=\log^2 2
\end{align}$$
